Personally, I find it difficult to interpret large areas. I prefer to find an approximate length to help imagine the scale. If you approximate the shape as a square, the calculation is pretty quick.
For example, maybe you are told that the lake below has an area of \(290\)\(\mathrm{miles^2}\).
If you approximate the shape as a square, what is the edge length? In other words, what is the approximate distance across the lake?
Solution
First, imagine the lake is a square with side lengths equaling \(s\).
The area of the square equals the side length raised to the power of two.
Personally, I find it difficult to interpret large areas. I prefer to find an approximate length to help imagine the scale. If you approximate the shape as a square, the calculation is pretty quick.
For example, maybe you are told that the lake below has an area of \(140\)\(\mathrm{miles^2}\).
If you approximate the shape as a square, what is the edge length? In other words, what is the approximate distance across the lake?
Solution
First, imagine the lake is a square with side lengths equaling \(s\).
The area of the square equals the side length raised to the power of two.
Personally, I find it difficult to interpret large areas. I prefer to find an approximate length to help imagine the scale. If you approximate the shape as a square, the calculation is pretty quick.
For example, maybe you are told that the lake below has an area of \(5900\)\(\mathrm{miles^2}\).
If you approximate the shape as a square, what is the edge length? In other words, what is the approximate distance across the lake?
Solution
First, imagine the lake is a square with side lengths equaling \(s\).
The area of the square equals the side length raised to the power of two.
Personally, I find it difficult to interpret large areas. I prefer to find an approximate length to help imagine the scale. If you approximate the shape as a square, the calculation is pretty quick.
For example, maybe you are told that the lake below has an area of \(150\)\(\mathrm{miles^2}\).
If you approximate the shape as a square, what is the edge length? In other words, what is the approximate distance across the lake?
Solution
First, imagine the lake is a square with side lengths equaling \(s\).
The area of the square equals the side length raised to the power of two.
Personally, I find it difficult to interpret large areas. I prefer to find an approximate length to help imagine the scale. If you approximate the shape as a square, the calculation is pretty quick.
For example, maybe you are told that the lake below has an area of \(2300\)\(\mathrm{miles^2}\).
If you approximate the shape as a square, what is the edge length? In other words, what is the approximate distance across the lake?
Solution
First, imagine the lake is a square with side lengths equaling \(s\).
The area of the square equals the side length raised to the power of two.
Personally, I find it difficult to interpret large areas. I prefer to find an approximate length to help imagine the scale. If you approximate the shape as a square, the calculation is pretty quick.
For example, maybe you are told that the lake below has an area of \(78\)\(\mathrm{miles^2}\).
If you approximate the shape as a square, what is the edge length? In other words, what is the approximate distance across the lake?
Solution
First, imagine the lake is a square with side lengths equaling \(s\).
The area of the square equals the side length raised to the power of two.
Personally, I find it difficult to interpret large areas. I prefer to find an approximate length to help imagine the scale. If you approximate the shape as a square, the calculation is pretty quick.
For example, maybe you are told that the lake below has an area of \(1500\)\(\mathrm{miles^2}\).
If you approximate the shape as a square, what is the edge length? In other words, what is the approximate distance across the lake?
Solution
First, imagine the lake is a square with side lengths equaling \(s\).
The area of the square equals the side length raised to the power of two.
Personally, I find it difficult to interpret large areas. I prefer to find an approximate length to help imagine the scale. If you approximate the shape as a square, the calculation is pretty quick.
For example, maybe you are told that the lake below has an area of \(760\)\(\mathrm{miles^2}\).
If you approximate the shape as a square, what is the edge length? In other words, what is the approximate distance across the lake?
Solution
First, imagine the lake is a square with side lengths equaling \(s\).
The area of the square equals the side length raised to the power of two.
Personally, I find it difficult to interpret large areas. I prefer to find an approximate length to help imagine the scale. If you approximate the shape as a square, the calculation is pretty quick.
For example, maybe you are told that the lake below has an area of \(250\)\(\mathrm{miles^2}\).
If you approximate the shape as a square, what is the edge length? In other words, what is the approximate distance across the lake?
Solution
First, imagine the lake is a square with side lengths equaling \(s\).
The area of the square equals the side length raised to the power of two.
Personally, I find it difficult to interpret large areas. I prefer to find an approximate length to help imagine the scale. If you approximate the shape as a square, the calculation is pretty quick.
For example, maybe you are told that the lake below has an area of \(4100\)\(\mathrm{miles^2}\).
If you approximate the shape as a square, what is the edge length? In other words, what is the approximate distance across the lake?
Solution
First, imagine the lake is a square with side lengths equaling \(s\).
The area of the square equals the side length raised to the power of two.
Consider the \(y=\sqrt{x}\) curve shown below (the square root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt{x}\)
22.3
5.51
54.5
8.38
74.6
9.04
Solution
To find \(y\), use square root or raising to the power of \(1/2\). To find \(x\), square (raise to the power of \(2\)).
\(x\)
\(y=\sqrt{x}\)
22.3
\(\sqrt{22.3}=(22.3)^{1/2}\approx4.72\)
\(5.51^2=30.4\)
5.51
54.5
\(\sqrt{54.5}=(54.5)^{1/2}\approx7.38\)
\(8.38^2=70.2\)
8.38
74.6
\(\sqrt{74.6}=(74.6)^{1/2}\approx8.64\)
\(9.04^2=81.7\)
9.04
Question
Consider the \(y=\sqrt{x}\) curve shown below (the square root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt{x}\)
9.5
5.5
37.8
6.78
58.2
9.75
Solution
To find \(y\), use square root or raising to the power of \(1/2\). To find \(x\), square (raise to the power of \(2\)).
\(x\)
\(y=\sqrt{x}\)
9.5
\(\sqrt{9.5}=(9.5)^{1/2}\approx3.08\)
\(5.5^2=30.2\)
5.5
37.8
\(\sqrt{37.8}=(37.8)^{1/2}\approx6.15\)
\(6.78^2=46\)
6.78
58.2
\(\sqrt{58.2}=(58.2)^{1/2}\approx7.63\)
\(9.75^2=95.1\)
9.75
Question
Consider the \(y=\sqrt{x}\) curve shown below (the square root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt{x}\)
25.1
5.7
37
6.7
76.7
9.09
Solution
To find \(y\), use square root or raising to the power of \(1/2\). To find \(x\), square (raise to the power of \(2\)).
\(x\)
\(y=\sqrt{x}\)
25.1
\(\sqrt{25.1}=(25.1)^{1/2}\approx5.01\)
\(5.7^2=32.5\)
5.7
37
\(\sqrt{37}=(37)^{1/2}\approx6.08\)
\(6.7^2=44.9\)
6.7
76.7
\(\sqrt{76.7}=(76.7)^{1/2}\approx8.76\)
\(9.09^2=82.6\)
9.09
Question
Consider the \(y=\sqrt{x}\) curve shown below (the square root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt{x}\)
8.6
3.38
13.3
4.76
36.8
9.06
Solution
To find \(y\), use square root or raising to the power of \(1/2\). To find \(x\), square (raise to the power of \(2\)).
\(x\)
\(y=\sqrt{x}\)
8.6
\(\sqrt{8.6}=(8.6)^{1/2}\approx2.93\)
\(3.38^2=11.4\)
3.38
13.3
\(\sqrt{13.3}=(13.3)^{1/2}\approx3.65\)
\(4.76^2=22.7\)
4.76
36.8
\(\sqrt{36.8}=(36.8)^{1/2}\approx6.07\)
\(9.06^2=82.1\)
9.06
Question
Consider the \(y=\sqrt{x}\) curve shown below (the square root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt{x}\)
5
3.69
65.4
8.56
79.4
9.41
Solution
To find \(y\), use square root or raising to the power of \(1/2\). To find \(x\), square (raise to the power of \(2\)).
\(x\)
\(y=\sqrt{x}\)
5
\(\sqrt{5}=(5)^{1/2}\approx2.24\)
\(3.69^2=13.6\)
3.69
65.4
\(\sqrt{65.4}=(65.4)^{1/2}\approx8.09\)
\(8.56^2=73.3\)
8.56
79.4
\(\sqrt{79.4}=(79.4)^{1/2}\approx8.91\)
\(9.41^2=88.5\)
9.41
Question
Consider the \(y=\sqrt{x}\) curve shown below (the square root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt{x}\)
8.2
3.61
24.3
7.02
54.8
7.96
Solution
To find \(y\), use square root or raising to the power of \(1/2\). To find \(x\), square (raise to the power of \(2\)).
\(x\)
\(y=\sqrt{x}\)
8.2
\(\sqrt{8.2}=(8.2)^{1/2}\approx2.86\)
\(3.61^2=13\)
3.61
24.3
\(\sqrt{24.3}=(24.3)^{1/2}\approx4.93\)
\(7.02^2=49.3\)
7.02
54.8
\(\sqrt{54.8}=(54.8)^{1/2}\approx7.4\)
\(7.96^2=63.4\)
7.96
Question
Consider the \(y=\sqrt{x}\) curve shown below (the square root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt{x}\)
4
5.8
55.8
8.21
75.2
8.81
Solution
To find \(y\), use square root or raising to the power of \(1/2\). To find \(x\), square (raise to the power of \(2\)).
\(x\)
\(y=\sqrt{x}\)
4
\(\sqrt{4}=(4)^{1/2}\approx2\)
\(5.8^2=33.6\)
5.8
55.8
\(\sqrt{55.8}=(55.8)^{1/2}\approx7.47\)
\(8.21^2=67.4\)
8.21
75.2
\(\sqrt{75.2}=(75.2)^{1/2}\approx8.67\)
\(8.81^2=77.6\)
8.81
Question
Consider the \(y=\sqrt{x}\) curve shown below (the square root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt{x}\)
4
5.26
29.3
5.86
58.1
9.85
Solution
To find \(y\), use square root or raising to the power of \(1/2\). To find \(x\), square (raise to the power of \(2\)).
\(x\)
\(y=\sqrt{x}\)
4
\(\sqrt{4}=(4)^{1/2}\approx2\)
\(5.26^2=27.7\)
5.26
29.3
\(\sqrt{29.3}=(29.3)^{1/2}\approx5.41\)
\(5.86^2=34.3\)
5.86
58.1
\(\sqrt{58.1}=(58.1)^{1/2}\approx7.62\)
\(9.85^2=97\)
9.85
Question
Consider the \(y=\sqrt{x}\) curve shown below (the square root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt{x}\)
4.2
3.08
12.6
4.75
44.6
8.08
Solution
To find \(y\), use square root or raising to the power of \(1/2\). To find \(x\), square (raise to the power of \(2\)).
\(x\)
\(y=\sqrt{x}\)
4.2
\(\sqrt{4.2}=(4.2)^{1/2}\approx2.05\)
\(3.08^2=9.5\)
3.08
12.6
\(\sqrt{12.6}=(12.6)^{1/2}\approx3.55\)
\(4.75^2=22.6\)
4.75
44.6
\(\sqrt{44.6}=(44.6)^{1/2}\approx6.68\)
\(8.08^2=65.3\)
8.08
Question
Consider the \(y=\sqrt{x}\) curve shown below (the square root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt{x}\)
19.4
5.77
41.6
6.95
72.9
8.6
Solution
To find \(y\), use square root or raising to the power of \(1/2\). To find \(x\), square (raise to the power of \(2\)).
\(x\)
\(y=\sqrt{x}\)
19.4
\(\sqrt{19.4}=(19.4)^{1/2}\approx4.4\)
\(5.77^2=33.3\)
5.77
41.6
\(\sqrt{41.6}=(41.6)^{1/2}\approx6.45\)
\(6.95^2=48.3\)
6.95
72.9
\(\sqrt{72.9}=(72.9)^{1/2}\approx8.54\)
\(8.6^2=74\)
8.6
Question
Consider the \(y=\sqrt{x}\) curve shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 8 graphs are made by translating and reflecting the \(y=\sqrt{x}\) curve. So each one is a different daughter function, made by altering the parent function.
Match the 8 graphs with the equations.
Equation \(y=-\sqrt{-(x+4)}-3\) matches graph .
Equation \(y=\sqrt{-(x-4)}+3\) matches graph .
Equation \(y=-\sqrt{-(x-4)}+3\) matches graph .
Equation \(y=\sqrt{x-4}-3\) matches graph .
Equation \(y=-\sqrt{-(x+4)}+3\) matches graph .
Equation \(y=-\sqrt{x-4}+3\) matches graph .
Equation \(y=\sqrt{-(x+4)}-3\) matches graph .
Equation \(y=\sqrt{x-4}+3\) matches graph .
Solution
Let’s call the origin of the parent function the “starting point” of the curve. It is useful to identify where the starting point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=-\sqrt{-(x+4)}-3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(4))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=4\]\[q=-3\]This tells us the new starting point is at \((-4,-3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph C.
Equation 2
The curve’s equation is \(y=\sqrt{-(x-4)}+3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(-4))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=-4\]\[q=3\]This tells us the new starting point is at \((4,3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph H.
Equation 3
The curve’s equation is \(y=-\sqrt{-(x-4)}+3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(-4))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=-4\]\[q=3\]This tells us the new starting point is at \((4,3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph A.
Equation 4
The curve’s equation is \(y=\sqrt{x-4}-3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(-4))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-4\]\[q=-3\]This tells us the new starting point is at \((4,-3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph D.
Equation 5
The curve’s equation is \(y=-\sqrt{-(x+4)}+3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(4))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=4\]\[q=3\]This tells us the new starting point is at \((-4,3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph G.
Equation 6
The curve’s equation is \(y=-\sqrt{x-4}+3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(-4))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=-4\]\[q=3\]This tells us the new starting point is at \((4,3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph F.
Equation 7
The curve’s equation is \(y=\sqrt{-(x+4)}-3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(4))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=4\]\[q=-3\]This tells us the new starting point is at \((-4,-3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph B.
Equation 8
The curve’s equation is \(y=\sqrt{x-4}+3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(-4))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-4\]\[q=3\]This tells us the new starting point is at \((4,3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph E.
Question
Consider the \(y=\sqrt{x}\) curve shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 8 graphs are made by translating and reflecting the \(y=\sqrt{x}\) curve. So each one is a different daughter function, made by altering the parent function.
Match the 8 graphs with the equations.
Equation \(y=-\sqrt{-(x-5)}-2\) matches graph .
Equation \(y=\sqrt{x-5}-2\) matches graph .
Equation \(y=-\sqrt{x-5}-2\) matches graph .
Equation \(y=\sqrt{-(x+5)}-2\) matches graph .
Equation \(y=-\sqrt{x+5}+2\) matches graph .
Equation \(y=-\sqrt{-(x-5)}+2\) matches graph .
Equation \(y=-\sqrt{-(x+5)}-2\) matches graph .
Equation \(y=-\sqrt{x+5}-2\) matches graph .
Solution
Let’s call the origin of the parent function the “starting point” of the curve. It is useful to identify where the starting point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=-\sqrt{-(x-5)}-2\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(-5))}+(-2)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=-5\]\[q=-2\]This tells us the new starting point is at \((5,-2)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph G.
Equation 2
The curve’s equation is \(y=\sqrt{x-5}-2\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(-5))}+(-2)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-5\]\[q=-2\]This tells us the new starting point is at \((5,-2)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph A.
Equation 3
The curve’s equation is \(y=-\sqrt{x-5}-2\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(-5))}+(-2)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=-5\]\[q=-2\]This tells us the new starting point is at \((5,-2)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph E.
Equation 4
The curve’s equation is \(y=\sqrt{-(x+5)}-2\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(5))}+(-2)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=5\]\[q=-2\]This tells us the new starting point is at \((-5,-2)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph C.
Equation 5
The curve’s equation is \(y=-\sqrt{x+5}+2\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(5))}+(2)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=5\]\[q=2\]This tells us the new starting point is at \((-5,2)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph F.
Equation 6
The curve’s equation is \(y=-\sqrt{-(x-5)}+2\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(-5))}+(2)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=-5\]\[q=2\]This tells us the new starting point is at \((5,2)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph H.
Equation 7
The curve’s equation is \(y=-\sqrt{-(x+5)}-2\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(5))}+(-2)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=5\]\[q=-2\]This tells us the new starting point is at \((-5,-2)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph B.
Equation 8
The curve’s equation is \(y=-\sqrt{x+5}-2\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(5))}+(-2)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=5\]\[q=-2\]This tells us the new starting point is at \((-5,-2)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph D.
Question
Consider the \(y=\sqrt{x}\) curve shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 8 graphs are made by translating and reflecting the \(y=\sqrt{x}\) curve. So each one is a different daughter function, made by altering the parent function.
Match the 8 graphs with the equations.
Equation \(y=-\sqrt{x-2}+5\) matches graph .
Equation \(y=\sqrt{-(x+2)}+5\) matches graph .
Equation \(y=-\sqrt{-(x-2)}-5\) matches graph .
Equation \(y=\sqrt{x+2}+5\) matches graph .
Equation \(y=-\sqrt{x+2}-5\) matches graph .
Equation \(y=\sqrt{x-2}-5\) matches graph .
Equation \(y=\sqrt{-(x-2)}+5\) matches graph .
Equation \(y=-\sqrt{-(x+2)}+5\) matches graph .
Solution
Let’s call the origin of the parent function the “starting point” of the curve. It is useful to identify where the starting point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=-\sqrt{x-2}+5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(-2))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=-2\]\[q=5\]This tells us the new starting point is at \((2,5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph D.
Equation 2
The curve’s equation is \(y=\sqrt{-(x+2)}+5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(2))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=2\]\[q=5\]This tells us the new starting point is at \((-2,5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph G.
Equation 3
The curve’s equation is \(y=-\sqrt{-(x-2)}-5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(-2))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=-2\]\[q=-5\]This tells us the new starting point is at \((2,-5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph E.
Equation 4
The curve’s equation is \(y=\sqrt{x+2}+5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(2))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=2\]\[q=5\]This tells us the new starting point is at \((-2,5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph B.
Equation 5
The curve’s equation is \(y=-\sqrt{x+2}-5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(2))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=2\]\[q=-5\]This tells us the new starting point is at \((-2,-5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph A.
Equation 6
The curve’s equation is \(y=\sqrt{x-2}-5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(-2))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-2\]\[q=-5\]This tells us the new starting point is at \((2,-5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph H.
Equation 7
The curve’s equation is \(y=\sqrt{-(x-2)}+5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(-2))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=-2\]\[q=5\]This tells us the new starting point is at \((2,5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph F.
Equation 8
The curve’s equation is \(y=-\sqrt{-(x+2)}+5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(2))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=2\]\[q=5\]This tells us the new starting point is at \((-2,5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph C.
Question
Consider the \(y=\sqrt{x}\) curve shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 8 graphs are made by translating and reflecting the \(y=\sqrt{x}\) curve. So each one is a different daughter function, made by altering the parent function.
Match the 8 graphs with the equations.
Equation \(y=\sqrt{x+2}+5\) matches graph .
Equation \(y=\sqrt{-(x+2)}+5\) matches graph .
Equation \(y=\sqrt{-(x-2)}-5\) matches graph .
Equation \(y=\sqrt{-(x+2)}-5\) matches graph .
Equation \(y=-\sqrt{x+2}+5\) matches graph .
Equation \(y=-\sqrt{x+2}-5\) matches graph .
Equation \(y=-\sqrt{x-2}-5\) matches graph .
Equation \(y=\sqrt{-(x-2)}+5\) matches graph .
Solution
Let’s call the origin of the parent function the “starting point” of the curve. It is useful to identify where the starting point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=\sqrt{x+2}+5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(2))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=2\]\[q=5\]This tells us the new starting point is at \((-2,5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph A.
Equation 2
The curve’s equation is \(y=\sqrt{-(x+2)}+5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(2))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=2\]\[q=5\]This tells us the new starting point is at \((-2,5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph C.
Equation 3
The curve’s equation is \(y=\sqrt{-(x-2)}-5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(-2))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=-2\]\[q=-5\]This tells us the new starting point is at \((2,-5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph H.
Equation 4
The curve’s equation is \(y=\sqrt{-(x+2)}-5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(2))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=2\]\[q=-5\]This tells us the new starting point is at \((-2,-5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph E.
Equation 5
The curve’s equation is \(y=-\sqrt{x+2}+5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(2))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=2\]\[q=5\]This tells us the new starting point is at \((-2,5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph B.
Equation 6
The curve’s equation is \(y=-\sqrt{x+2}-5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(2))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=2\]\[q=-5\]This tells us the new starting point is at \((-2,-5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph G.
Equation 7
The curve’s equation is \(y=-\sqrt{x-2}-5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(-2))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=-2\]\[q=-5\]This tells us the new starting point is at \((2,-5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph D.
Equation 8
The curve’s equation is \(y=\sqrt{-(x-2)}+5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(-2))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=-2\]\[q=5\]This tells us the new starting point is at \((2,5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph F.
Question
Consider the \(y=\sqrt{x}\) curve shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 8 graphs are made by translating and reflecting the \(y=\sqrt{x}\) curve. So each one is a different daughter function, made by altering the parent function.
Match the 8 graphs with the equations.
Equation \(y=\sqrt{x-5}-3\) matches graph .
Equation \(y=-\sqrt{-(x-5)}+3\) matches graph .
Equation \(y=\sqrt{-(x+5)}-3\) matches graph .
Equation \(y=-\sqrt{x+5}+3\) matches graph .
Equation \(y=-\sqrt{x-5}+3\) matches graph .
Equation \(y=\sqrt{x+5}+3\) matches graph .
Equation \(y=\sqrt{-(x+5)}+3\) matches graph .
Equation \(y=\sqrt{-(x-5)}+3\) matches graph .
Solution
Let’s call the origin of the parent function the “starting point” of the curve. It is useful to identify where the starting point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=\sqrt{x-5}-3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(-5))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-5\]\[q=-3\]This tells us the new starting point is at \((5,-3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph E.
Equation 2
The curve’s equation is \(y=-\sqrt{-(x-5)}+3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(-5))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=-5\]\[q=3\]This tells us the new starting point is at \((5,3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph D.
Equation 3
The curve’s equation is \(y=\sqrt{-(x+5)}-3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(5))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=5\]\[q=-3\]This tells us the new starting point is at \((-5,-3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph G.
Equation 4
The curve’s equation is \(y=-\sqrt{x+5}+3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(5))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=5\]\[q=3\]This tells us the new starting point is at \((-5,3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph B.
Equation 5
The curve’s equation is \(y=-\sqrt{x-5}+3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(-5))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=-5\]\[q=3\]This tells us the new starting point is at \((5,3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph A.
Equation 6
The curve’s equation is \(y=\sqrt{x+5}+3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(5))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=5\]\[q=3\]This tells us the new starting point is at \((-5,3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph F.
Equation 7
The curve’s equation is \(y=\sqrt{-(x+5)}+3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(5))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=5\]\[q=3\]This tells us the new starting point is at \((-5,3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph H.
Equation 8
The curve’s equation is \(y=\sqrt{-(x-5)}+3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(-5))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=-5\]\[q=3\]This tells us the new starting point is at \((5,3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph C.
Question
Consider the \(y=\sqrt{x}\) curve shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 8 graphs are made by translating and reflecting the \(y=\sqrt{x}\) curve. So each one is a different daughter function, made by altering the parent function.
Match the 8 graphs with the equations.
Equation \(y=\sqrt{-(x+5)}-3\) matches graph .
Equation \(y=-\sqrt{-(x+5)}+3\) matches graph .
Equation \(y=-\sqrt{-(x-5)}-3\) matches graph .
Equation \(y=-\sqrt{x+5}+3\) matches graph .
Equation \(y=\sqrt{x+5}+3\) matches graph .
Equation \(y=\sqrt{x-5}+3\) matches graph .
Equation \(y=-\sqrt{-(x+5)}-3\) matches graph .
Equation \(y=\sqrt{-(x-5)}-3\) matches graph .
Solution
Let’s call the origin of the parent function the “starting point” of the curve. It is useful to identify where the starting point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=\sqrt{-(x+5)}-3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(5))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=5\]\[q=-3\]This tells us the new starting point is at \((-5,-3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph A.
Equation 2
The curve’s equation is \(y=-\sqrt{-(x+5)}+3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(5))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=5\]\[q=3\]This tells us the new starting point is at \((-5,3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph H.
Equation 3
The curve’s equation is \(y=-\sqrt{-(x-5)}-3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(-5))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=-5\]\[q=-3\]This tells us the new starting point is at \((5,-3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph F.
Equation 4
The curve’s equation is \(y=-\sqrt{x+5}+3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(5))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=5\]\[q=3\]This tells us the new starting point is at \((-5,3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph B.
Equation 5
The curve’s equation is \(y=\sqrt{x+5}+3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(5))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=5\]\[q=3\]This tells us the new starting point is at \((-5,3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph C.
Equation 6
The curve’s equation is \(y=\sqrt{x-5}+3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(-5))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-5\]\[q=3\]This tells us the new starting point is at \((5,3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph D.
Equation 7
The curve’s equation is \(y=-\sqrt{-(x+5)}-3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(5))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=5\]\[q=-3\]This tells us the new starting point is at \((-5,-3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph E.
Equation 8
The curve’s equation is \(y=\sqrt{-(x-5)}-3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(-5))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=-5\]\[q=-3\]This tells us the new starting point is at \((5,-3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph G.
Question
Consider the \(y=\sqrt{x}\) curve shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 8 graphs are made by translating and reflecting the \(y=\sqrt{x}\) curve. So each one is a different daughter function, made by altering the parent function.
Match the 8 graphs with the equations.
Equation \(y=\sqrt{x+4}-3\) matches graph .
Equation \(y=\sqrt{x+4}+3\) matches graph .
Equation \(y=-\sqrt{x+4}-3\) matches graph .
Equation \(y=-\sqrt{x-4}-3\) matches graph .
Equation \(y=\sqrt{-(x+4)}-3\) matches graph .
Equation \(y=\sqrt{-(x-4)}-3\) matches graph .
Equation \(y=-\sqrt{-(x+4)}+3\) matches graph .
Equation \(y=\sqrt{x-4}+3\) matches graph .
Solution
Let’s call the origin of the parent function the “starting point” of the curve. It is useful to identify where the starting point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=\sqrt{x+4}-3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(4))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=4\]\[q=-3\]This tells us the new starting point is at \((-4,-3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph E.
Equation 2
The curve’s equation is \(y=\sqrt{x+4}+3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(4))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=4\]\[q=3\]This tells us the new starting point is at \((-4,3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph H.
Equation 3
The curve’s equation is \(y=-\sqrt{x+4}-3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(4))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=4\]\[q=-3\]This tells us the new starting point is at \((-4,-3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph A.
Equation 4
The curve’s equation is \(y=-\sqrt{x-4}-3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(-4))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=-4\]\[q=-3\]This tells us the new starting point is at \((4,-3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph F.
Equation 5
The curve’s equation is \(y=\sqrt{-(x+4)}-3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(4))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=4\]\[q=-3\]This tells us the new starting point is at \((-4,-3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph B.
Equation 6
The curve’s equation is \(y=\sqrt{-(x-4)}-3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(-4))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=-4\]\[q=-3\]This tells us the new starting point is at \((4,-3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph G.
Equation 7
The curve’s equation is \(y=-\sqrt{-(x+4)}+3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(4))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=4\]\[q=3\]This tells us the new starting point is at \((-4,3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph D.
Equation 8
The curve’s equation is \(y=\sqrt{x-4}+3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(-4))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-4\]\[q=3\]This tells us the new starting point is at \((4,3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph C.
Question
Consider the \(y=\sqrt{x}\) curve shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 8 graphs are made by translating and reflecting the \(y=\sqrt{x}\) curve. So each one is a different daughter function, made by altering the parent function.
Match the 8 graphs with the equations.
Equation \(y=-\sqrt{-(x+4)}-2\) matches graph .
Equation \(y=\sqrt{-(x-4)}+2\) matches graph .
Equation \(y=-\sqrt{-(x-4)}+2\) matches graph .
Equation \(y=\sqrt{-(x+4)}-2\) matches graph .
Equation \(y=-\sqrt{x-4}+2\) matches graph .
Equation \(y=\sqrt{x-4}-2\) matches graph .
Equation \(y=-\sqrt{x-4}-2\) matches graph .
Equation \(y=\sqrt{x+4}-2\) matches graph .
Solution
Let’s call the origin of the parent function the “starting point” of the curve. It is useful to identify where the starting point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=-\sqrt{-(x+4)}-2\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(4))}+(-2)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=4\]\[q=-2\]This tells us the new starting point is at \((-4,-2)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph B.
Equation 2
The curve’s equation is \(y=\sqrt{-(x-4)}+2\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(-4))}+(2)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=-4\]\[q=2\]This tells us the new starting point is at \((4,2)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph E.
Equation 3
The curve’s equation is \(y=-\sqrt{-(x-4)}+2\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(-4))}+(2)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=-4\]\[q=2\]This tells us the new starting point is at \((4,2)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph A.
Equation 4
The curve’s equation is \(y=\sqrt{-(x+4)}-2\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(4))}+(-2)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=4\]\[q=-2\]This tells us the new starting point is at \((-4,-2)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph H.
Equation 5
The curve’s equation is \(y=-\sqrt{x-4}+2\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(-4))}+(2)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=-4\]\[q=2\]This tells us the new starting point is at \((4,2)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph D.
Equation 6
The curve’s equation is \(y=\sqrt{x-4}-2\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(-4))}+(-2)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-4\]\[q=-2\]This tells us the new starting point is at \((4,-2)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph F.
Equation 7
The curve’s equation is \(y=-\sqrt{x-4}-2\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(-4))}+(-2)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=-4\]\[q=-2\]This tells us the new starting point is at \((4,-2)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph G.
Equation 8
The curve’s equation is \(y=\sqrt{x+4}-2\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(4))}+(-2)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=4\]\[q=-2\]This tells us the new starting point is at \((-4,-2)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph C.
Question
Consider the \(y=\sqrt{x}\) curve shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 8 graphs are made by translating and reflecting the \(y=\sqrt{x}\) curve. So each one is a different daughter function, made by altering the parent function.
Match the 8 graphs with the equations.
Equation \(y=\sqrt{x+3}+5\) matches graph .
Equation \(y=-\sqrt{-(x+3)}-5\) matches graph .
Equation \(y=-\sqrt{-(x+3)}+5\) matches graph .
Equation \(y=-\sqrt{-(x-3)}+5\) matches graph .
Equation \(y=-\sqrt{x+3}+5\) matches graph .
Equation \(y=\sqrt{-(x-3)}-5\) matches graph .
Equation \(y=-\sqrt{x+3}-5\) matches graph .
Equation \(y=\sqrt{x+3}-5\) matches graph .
Solution
Let’s call the origin of the parent function the “starting point” of the curve. It is useful to identify where the starting point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=\sqrt{x+3}+5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(3))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=3\]\[q=5\]This tells us the new starting point is at \((-3,5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph D.
Equation 2
The curve’s equation is \(y=-\sqrt{-(x+3)}-5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(3))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=3\]\[q=-5\]This tells us the new starting point is at \((-3,-5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph H.
Equation 3
The curve’s equation is \(y=-\sqrt{-(x+3)}+5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(3))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=3\]\[q=5\]This tells us the new starting point is at \((-3,5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph A.
Equation 4
The curve’s equation is \(y=-\sqrt{-(x-3)}+5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(-3))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=-3\]\[q=5\]This tells us the new starting point is at \((3,5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph G.
Equation 5
The curve’s equation is \(y=-\sqrt{x+3}+5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(3))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=3\]\[q=5\]This tells us the new starting point is at \((-3,5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph C.
Equation 6
The curve’s equation is \(y=\sqrt{-(x-3)}-5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(-3))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=-3\]\[q=-5\]This tells us the new starting point is at \((3,-5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph F.
Equation 7
The curve’s equation is \(y=-\sqrt{x+3}-5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(3))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=3\]\[q=-5\]This tells us the new starting point is at \((-3,-5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph B.
Equation 8
The curve’s equation is \(y=\sqrt{x+3}-5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(3))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=3\]\[q=-5\]This tells us the new starting point is at \((-3,-5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph E.
Question
Consider the \(y=\sqrt{x}\) curve shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 8 graphs are made by translating and reflecting the \(y=\sqrt{x}\) curve. So each one is a different daughter function, made by altering the parent function.
Match the 8 graphs with the equations.
Equation \(y=\sqrt{x+3}-5\) matches graph .
Equation \(y=-\sqrt{-(x-3)}+5\) matches graph .
Equation \(y=\sqrt{-(x-3)}+5\) matches graph .
Equation \(y=-\sqrt{x-3}-5\) matches graph .
Equation \(y=\sqrt{x-3}-5\) matches graph .
Equation \(y=-\sqrt{-(x+3)}+5\) matches graph .
Equation \(y=\sqrt{-(x-3)}-5\) matches graph .
Equation \(y=-\sqrt{x+3}+5\) matches graph .
Solution
Let’s call the origin of the parent function the “starting point” of the curve. It is useful to identify where the starting point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=\sqrt{x+3}-5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(3))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=3\]\[q=-5\]This tells us the new starting point is at \((-3,-5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph D.
Equation 2
The curve’s equation is \(y=-\sqrt{-(x-3)}+5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(-3))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=-3\]\[q=5\]This tells us the new starting point is at \((3,5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph A.
Equation 3
The curve’s equation is \(y=\sqrt{-(x-3)}+5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(-3))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=-3\]\[q=5\]This tells us the new starting point is at \((3,5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph B.
Equation 4
The curve’s equation is \(y=-\sqrt{x-3}-5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(-3))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=-3\]\[q=-5\]This tells us the new starting point is at \((3,-5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph F.
Equation 5
The curve’s equation is \(y=\sqrt{x-3}-5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(1)(x+(-3))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-3\]\[q=-5\]This tells us the new starting point is at \((3,-5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and up. Thus, the correct graph is graph C.
Equation 6
The curve’s equation is \(y=-\sqrt{-(x+3)}+5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(-1)(x+(3))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=3\]\[q=5\]This tells us the new starting point is at \((-3,5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and down. Thus, the correct graph is graph E.
Equation 7
The curve’s equation is \(y=\sqrt{-(x-3)}-5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt{(-1)(x+(-3))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=-3\]\[q=-5\]This tells us the new starting point is at \((3,-5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). This means, from the starting point, the curve heads left and up. Thus, the correct graph is graph G.
Equation 8
The curve’s equation is \(y=-\sqrt{x+3}+5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt{(1)(x+(3))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=3\]\[q=5\]This tells us the new starting point is at \((-3,5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). This means, from the starting point, the curve heads right and down. Thus, the correct graph is graph H.
Question
The function \(f(x)~=~\sqrt{x}\) has some integer input-output pairs (see Diophantine equations). The first 7 are listed and graphed below.
\(x\)
\(y=f(x)\)
0
0
1
1
4
2
9
3
16
4
25
5
36
6
A shifted function \(g\) is defined as \(g(x)=\sqrt{x-4}+2\). Find the corresponding points of integer input and integer output on function \(g\).
\(x\)
\(y=g(x)\)
Solution
Function \(g\) can be thought of a shifted version of \(f\), shifted horizontally by 4 and vertically by 2.
Graph of function \(g\)
\(x\)
\(y=f(x)\)
4
2
5
3
8
4
13
5
20
6
29
7
40
8
Question
The function \(f(x)~=~\sqrt{x}\) has some integer input-output pairs (see Diophantine equations). The first 7 are listed and graphed below.
\(x\)
\(y=f(x)\)
0
0
1
1
4
2
9
3
16
4
25
5
36
6
A shifted function \(g\) is defined as \(g(x)=\sqrt{x-5}+1\). Find the corresponding points of integer input and integer output on function \(g\).
\(x\)
\(y=g(x)\)
Solution
Function \(g\) can be thought of a shifted version of \(f\), shifted horizontally by 5 and vertically by 1.
Graph of function \(g\)
\(x\)
\(y=f(x)\)
5
1
6
2
9
3
14
4
21
5
30
6
41
7
Question
The function \(f(x)~=~\sqrt{x}\) has some integer input-output pairs (see Diophantine equations). The first 7 are listed and graphed below.
\(x\)
\(y=f(x)\)
0
0
1
1
4
2
9
3
16
4
25
5
36
6
A shifted function \(g\) is defined as \(g(x)=\sqrt{x-2}+4\). Find the corresponding points of integer input and integer output on function \(g\).
\(x\)
\(y=g(x)\)
Solution
Function \(g\) can be thought of a shifted version of \(f\), shifted horizontally by 2 and vertically by 4.
Graph of function \(g\)
\(x\)
\(y=f(x)\)
2
4
3
5
6
6
11
7
18
8
27
9
38
10
Question
The function \(f(x)~=~\sqrt{x}\) has some integer input-output pairs (see Diophantine equations). The first 7 are listed and graphed below.
\(x\)
\(y=f(x)\)
0
0
1
1
4
2
9
3
16
4
25
5
36
6
A shifted function \(g\) is defined as \(g(x)=\sqrt{x-5}+1\). Find the corresponding points of integer input and integer output on function \(g\).
\(x\)
\(y=g(x)\)
Solution
Function \(g\) can be thought of a shifted version of \(f\), shifted horizontally by 5 and vertically by 1.
Graph of function \(g\)
\(x\)
\(y=f(x)\)
5
1
6
2
9
3
14
4
21
5
30
6
41
7
Question
The function \(f(x)~=~\sqrt{x}\) has some integer input-output pairs (see Diophantine equations). The first 7 are listed and graphed below.
\(x\)
\(y=f(x)\)
0
0
1
1
4
2
9
3
16
4
25
5
36
6
A shifted function \(g\) is defined as \(g(x)=\sqrt{x+5}+4\). Find the corresponding points of integer input and integer output on function \(g\).
\(x\)
\(y=g(x)\)
Solution
Function \(g\) can be thought of a shifted version of \(f\), shifted horizontally by -5 and vertically by 4.
Graph of function \(g\)
\(x\)
\(y=f(x)\)
-5
4
-4
5
-1
6
4
7
11
8
20
9
31
10
Question
The function \(f(x)~=~\sqrt{x}\) has some integer input-output pairs (see Diophantine equations). The first 7 are listed and graphed below.
\(x\)
\(y=f(x)\)
0
0
1
1
4
2
9
3
16
4
25
5
36
6
A shifted function \(g\) is defined as \(g(x)=\sqrt{x+5}-2\). Find the corresponding points of integer input and integer output on function \(g\).
\(x\)
\(y=g(x)\)
Solution
Function \(g\) can be thought of a shifted version of \(f\), shifted horizontally by -5 and vertically by -2.
Graph of function \(g\)
\(x\)
\(y=f(x)\)
-5
-2
-4
-1
-1
0
4
1
11
2
20
3
31
4
Question
The function \(f(x)~=~\sqrt{x}\) has some integer input-output pairs (see Diophantine equations). The first 7 are listed and graphed below.
\(x\)
\(y=f(x)\)
0
0
1
1
4
2
9
3
16
4
25
5
36
6
A shifted function \(g\) is defined as \(g(x)=\sqrt{x-3}-5\). Find the corresponding points of integer input and integer output on function \(g\).
\(x\)
\(y=g(x)\)
Solution
Function \(g\) can be thought of a shifted version of \(f\), shifted horizontally by 3 and vertically by -5.
Graph of function \(g\)
\(x\)
\(y=f(x)\)
3
-5
4
-4
7
-3
12
-2
19
-1
28
0
39
1
Question
The function \(f(x)~=~\sqrt{x}\) has some integer input-output pairs (see Diophantine equations). The first 7 are listed and graphed below.
\(x\)
\(y=f(x)\)
0
0
1
1
4
2
9
3
16
4
25
5
36
6
A shifted function \(g\) is defined as \(g(x)=\sqrt{x+5}-1\). Find the corresponding points of integer input and integer output on function \(g\).
\(x\)
\(y=g(x)\)
Solution
Function \(g\) can be thought of a shifted version of \(f\), shifted horizontally by -5 and vertically by -1.
Graph of function \(g\)
\(x\)
\(y=f(x)\)
-5
-1
-4
0
-1
1
4
2
11
3
20
4
31
5
Question
The function \(f(x)~=~\sqrt{x}\) has some integer input-output pairs (see Diophantine equations). The first 7 are listed and graphed below.
\(x\)
\(y=f(x)\)
0
0
1
1
4
2
9
3
16
4
25
5
36
6
A shifted function \(g\) is defined as \(g(x)=\sqrt{x-1}-3\). Find the corresponding points of integer input and integer output on function \(g\).
\(x\)
\(y=g(x)\)
Solution
Function \(g\) can be thought of a shifted version of \(f\), shifted horizontally by 1 and vertically by -3.
Graph of function \(g\)
\(x\)
\(y=f(x)\)
1
-3
2
-2
5
-1
10
0
17
1
26
2
37
3
Question
The function \(f(x)~=~\sqrt{x}\) has some integer input-output pairs (see Diophantine equations). The first 7 are listed and graphed below.
\(x\)
\(y=f(x)\)
0
0
1
1
4
2
9
3
16
4
25
5
36
6
A shifted function \(g\) is defined as \(g(x)=\sqrt{x-3}-4\). Find the corresponding points of integer input and integer output on function \(g\).
\(x\)
\(y=g(x)\)
Solution
Function \(g\) can be thought of a shifted version of \(f\), shifted horizontally by 3 and vertically by -4.
Graph of function \(g\)
\(x\)
\(y=f(x)\)
3
-4
4
-3
7
-2
12
-1
19
0
28
1
39
2
Question
Consider the \(y=\sqrt[3]{x}\) curve shown below (the cube root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt[3]{x}\)
-899
-9.09
-66
4.56
599
9.33
Solution
To find \(y\), use cube root or raising to the power of \(1/3\). To find \(x\), cube (raise to the power of \(3\)).
\(x\)
\(y=\sqrt[3]{x}\)
-899
\(\sqrt[3]{-899}=(-899)^{1/3}\approx-9.65\)
\((-9.09)^3=-751\)
-9.09
-66
\(\sqrt[3]{-66}=(-66)^{1/3}\approx-4.04\)
\((4.56)^3=95\)
4.56
599
\(\sqrt[3]{599}=(599)^{1/3}\approx8.43\)
\((9.33)^3=812\)
9.33
Question
Consider the \(y=\sqrt[3]{x}\) curve shown below (the cube root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt[3]{x}\)
-825
-9.08
-355
-4.82
-35
4.06
Solution
To find \(y\), use cube root or raising to the power of \(1/3\). To find \(x\), cube (raise to the power of \(3\)).
\(x\)
\(y=\sqrt[3]{x}\)
-825
\(\sqrt[3]{-825}=(-825)^{1/3}\approx-9.38\)
\((-9.08)^3=-749\)
-9.08
-355
\(\sqrt[3]{-355}=(-355)^{1/3}\approx-7.08\)
\((-4.82)^3=-112\)
-4.82
-35
\(\sqrt[3]{-35}=(-35)^{1/3}\approx-3.27\)
\((4.06)^3=67\)
4.06
Question
Consider the \(y=\sqrt[3]{x}\) curve shown below (the cube root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt[3]{x}\)
-375
-6.6
-141
4.14
161
8.06
Solution
To find \(y\), use cube root or raising to the power of \(1/3\). To find \(x\), cube (raise to the power of \(3\)).
\(x\)
\(y=\sqrt[3]{x}\)
-375
\(\sqrt[3]{-375}=(-375)^{1/3}\approx-7.21\)
\((-6.6)^3=-287\)
-6.6
-141
\(\sqrt[3]{-141}=(-141)^{1/3}\approx-5.2\)
\((4.14)^3=71\)
4.14
161
\(\sqrt[3]{161}=(161)^{1/3}\approx5.44\)
\((8.06)^3=524\)
8.06
Question
Consider the \(y=\sqrt[3]{x}\) curve shown below (the cube root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt[3]{x}\)
-769
-7.67
-321
3.39
432
8.68
Solution
To find \(y\), use cube root or raising to the power of \(1/3\). To find \(x\), cube (raise to the power of \(3\)).
\(x\)
\(y=\sqrt[3]{x}\)
-769
\(\sqrt[3]{-769}=(-769)^{1/3}\approx-9.16\)
\((-7.67)^3=-451\)
-7.67
-321
\(\sqrt[3]{-321}=(-321)^{1/3}\approx-6.85\)
\((3.39)^3=39\)
3.39
432
\(\sqrt[3]{432}=(432)^{1/3}\approx7.56\)
\((8.68)^3=654\)
8.68
Question
Consider the \(y=\sqrt[3]{x}\) curve shown below (the cube root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt[3]{x}\)
-910
-8.48
-439
-5.96
514
8.54
Solution
To find \(y\), use cube root or raising to the power of \(1/3\). To find \(x\), cube (raise to the power of \(3\)).
\(x\)
\(y=\sqrt[3]{x}\)
-910
\(\sqrt[3]{-910}=(-910)^{1/3}\approx-9.69\)
\((-8.48)^3=-610\)
-8.48
-439
\(\sqrt[3]{-439}=(-439)^{1/3}\approx-7.6\)
\((-5.96)^3=-212\)
-5.96
514
\(\sqrt[3]{514}=(514)^{1/3}\approx8.01\)
\((8.54)^3=623\)
8.54
Question
Consider the \(y=\sqrt[3]{x}\) curve shown below (the cube root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt[3]{x}\)
-510
-7.19
-186
-4.85
432
8.85
Solution
To find \(y\), use cube root or raising to the power of \(1/3\). To find \(x\), cube (raise to the power of \(3\)).
\(x\)
\(y=\sqrt[3]{x}\)
-510
\(\sqrt[3]{-510}=(-510)^{1/3}\approx-7.99\)
\((-7.19)^3=-372\)
-7.19
-186
\(\sqrt[3]{-186}=(-186)^{1/3}\approx-5.71\)
\((-4.85)^3=-114\)
-4.85
432
\(\sqrt[3]{432}=(432)^{1/3}\approx7.56\)
\((8.85)^3=693\)
8.85
Question
Consider the \(y=\sqrt[3]{x}\) curve shown below (the cube root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt[3]{x}\)
-863
-9.15
243
6.87
469
8.6
Solution
To find \(y\), use cube root or raising to the power of \(1/3\). To find \(x\), cube (raise to the power of \(3\)).
\(x\)
\(y=\sqrt[3]{x}\)
-863
\(\sqrt[3]{-863}=(-863)^{1/3}\approx-9.52\)
\((-9.15)^3=-766\)
-9.15
243
\(\sqrt[3]{243}=(243)^{1/3}\approx6.24\)
\((6.87)^3=324\)
6.87
469
\(\sqrt[3]{469}=(469)^{1/3}\approx7.77\)
\((8.6)^3=636\)
8.6
Question
Consider the \(y=\sqrt[3]{x}\) curve shown below (the cube root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt[3]{x}\)
-555
-7.7
8
5.25
340
9.65
Solution
To find \(y\), use cube root or raising to the power of \(1/3\). To find \(x\), cube (raise to the power of \(3\)).
\(x\)
\(y=\sqrt[3]{x}\)
-555
\(\sqrt[3]{-555}=(-555)^{1/3}\approx-8.22\)
\((-7.7)^3=-457\)
-7.7
8
\(\sqrt[3]{8}=(8)^{1/3}\approx2\)
\((5.25)^3=145\)
5.25
340
\(\sqrt[3]{340}=(340)^{1/3}\approx6.98\)
\((9.65)^3=899\)
9.65
Question
Consider the \(y=\sqrt[3]{x}\) curve shown below (the cube root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt[3]{x}\)
-815
-5.68
97
5.73
681
9.68
Solution
To find \(y\), use cube root or raising to the power of \(1/3\). To find \(x\), cube (raise to the power of \(3\)).
\(x\)
\(y=\sqrt[3]{x}\)
-815
\(\sqrt[3]{-815}=(-815)^{1/3}\approx-9.34\)
\((-5.68)^3=-183\)
-5.68
97
\(\sqrt[3]{97}=(97)^{1/3}\approx4.59\)
\((5.73)^3=188\)
5.73
681
\(\sqrt[3]{681}=(681)^{1/3}\approx8.8\)
\((9.68)^3=907\)
9.68
Question
Consider the \(y=\sqrt[3]{x}\) curve shown below (the cube root function). There are 6 points along the curve, find the missing coordinates of those points in the table.
\(x\)
\(y=\sqrt[3]{x}\)
-512
-7.55
-323
-5.38
564
8.54
Solution
To find \(y\), use cube root or raising to the power of \(1/3\). To find \(x\), cube (raise to the power of \(3\)).
\(x\)
\(y=\sqrt[3]{x}\)
-512
\(\sqrt[3]{-512}=(-512)^{1/3}\approx-8\)
\((-7.55)^3=-430\)
-7.55
-323
\(\sqrt[3]{-323}=(-323)^{1/3}\approx-6.86\)
\((-5.38)^3=-156\)
-5.38
564
\(\sqrt[3]{564}=(564)^{1/3}\approx8.26\)
\((8.54)^3=623\)
8.54
Question
Consider the \(y=\sqrt[3]{x}\) curve (cube-root function) shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 4 graphs are made by translating and reflecting the \(y=\sqrt[3]{x}\) curve. So each one is a different daughter function, made by altering the parent function.
The origin of the parent cube-root function is called the “inflection point”. It is useful to identify where the inflection point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt[3]{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=-\sqrt[3]{-(x+5)}-4\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(-1)(x+(5))}+(-4)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=5\]\[q=-4\]This tells us the new inflection point is at \((-5,-4)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting both horizontally and vertically results in a return to the original increasing orientation. Thus, the correct graph is graph D.
Equation 2
The curve’s equation is \(y=-\sqrt[3]{x-5}+4\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(1)(x+(-5))}+(4)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=-5\]\[q=4\]This tells us the new inflection point is at \((5,4)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph B.
Equation 3
The curve’s equation is \(y=-\sqrt[3]{x+5}+4\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(1)(x+(5))}+(4)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=5\]\[q=4\]This tells us the new inflection point is at \((-5,4)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph A.
Equation 4
The curve’s equation is \(y=-\sqrt[3]{-(x+5)}+4\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(-1)(x+(5))}+(4)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=5\]\[q=4\]This tells us the new inflection point is at \((-5,4)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting both horizontally and vertically results in a return to the original increasing orientation. Thus, the correct graph is graph C.
Question
Consider the \(y=\sqrt[3]{x}\) curve (cube-root function) shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 4 graphs are made by translating and reflecting the \(y=\sqrt[3]{x}\) curve. So each one is a different daughter function, made by altering the parent function.
Match the 4 graphs with the equations.
Equation \(y=\sqrt[3]{x-4}+5\) matches graph .
Equation \(y=-\sqrt[3]{x-4}-5\) matches graph .
Equation \(y=\sqrt[3]{x-4}-5\) matches graph .
Equation \(y=\sqrt[3]{-(x-4)}+5\) matches graph .
Solution
The origin of the parent cube-root function is called the “inflection point”. It is useful to identify where the inflection point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt[3]{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=\sqrt[3]{x-4}+5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(1)(x+(-4))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-4\]\[q=5\]This tells us the new inflection point is at \((4,5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). Because there are no reflections, the curve is in the original increasing orientation. Thus, the correct graph is graph A.
Equation 2
The curve’s equation is \(y=-\sqrt[3]{x-4}-5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(1)(x+(-4))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=-4\]\[q=-5\]This tells us the new inflection point is at \((4,-5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph D.
Equation 3
The curve’s equation is \(y=\sqrt[3]{x-4}-5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(1)(x+(-4))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-4\]\[q=-5\]This tells us the new inflection point is at \((4,-5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). Because there are no reflections, the curve is in the original increasing orientation. Thus, the correct graph is graph C.
Equation 4
The curve’s equation is \(y=\sqrt[3]{-(x-4)}+5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(-1)(x+(-4))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=-4\]\[q=5\]This tells us the new inflection point is at \((4,5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph B.
Question
Consider the \(y=\sqrt[3]{x}\) curve (cube-root function) shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 4 graphs are made by translating and reflecting the \(y=\sqrt[3]{x}\) curve. So each one is a different daughter function, made by altering the parent function.
The origin of the parent cube-root function is called the “inflection point”. It is useful to identify where the inflection point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt[3]{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=\sqrt[3]{-(x+3)}-5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(-1)(x+(3))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=3\]\[q=-5\]This tells us the new inflection point is at \((-3,-5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph D.
Equation 2
The curve’s equation is \(y=\sqrt[3]{-(x-3)}-5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(-1)(x+(-3))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=-3\]\[q=-5\]This tells us the new inflection point is at \((3,-5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph B.
Equation 3
The curve’s equation is \(y=-\sqrt[3]{-(x+3)}-5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(-1)(x+(3))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=3\]\[q=-5\]This tells us the new inflection point is at \((-3,-5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting both horizontally and vertically results in a return to the original increasing orientation. Thus, the correct graph is graph C.
Equation 4
The curve’s equation is \(y=\sqrt[3]{x-3}-5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(1)(x+(-3))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-3\]\[q=-5\]This tells us the new inflection point is at \((3,-5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). Because there are no reflections, the curve is in the original increasing orientation. Thus, the correct graph is graph A.
Question
Consider the \(y=\sqrt[3]{x}\) curve (cube-root function) shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 4 graphs are made by translating and reflecting the \(y=\sqrt[3]{x}\) curve. So each one is a different daughter function, made by altering the parent function.
The origin of the parent cube-root function is called the “inflection point”. It is useful to identify where the inflection point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt[3]{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=-\sqrt[3]{x+5}-2\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(1)(x+(5))}+(-2)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=5\]\[q=-2\]This tells us the new inflection point is at \((-5,-2)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph A.
Equation 2
The curve’s equation is \(y=\sqrt[3]{-(x+5)}+2\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(-1)(x+(5))}+(2)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=5\]\[q=2\]This tells us the new inflection point is at \((-5,2)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph B.
Equation 3
The curve’s equation is \(y=-\sqrt[3]{-(x-5)}+2\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(-1)(x+(-5))}+(2)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=-5\]\[q=2\]This tells us the new inflection point is at \((5,2)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting both horizontally and vertically results in a return to the original increasing orientation. Thus, the correct graph is graph C.
Equation 4
The curve’s equation is \(y=\sqrt[3]{x+5}+2\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(1)(x+(5))}+(2)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=5\]\[q=2\]This tells us the new inflection point is at \((-5,2)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). Because there are no reflections, the curve is in the original increasing orientation. Thus, the correct graph is graph D.
Question
Consider the \(y=\sqrt[3]{x}\) curve (cube-root function) shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 4 graphs are made by translating and reflecting the \(y=\sqrt[3]{x}\) curve. So each one is a different daughter function, made by altering the parent function.
The origin of the parent cube-root function is called the “inflection point”. It is useful to identify where the inflection point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt[3]{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=-\sqrt[3]{x-3}-2\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(1)(x+(-3))}+(-2)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=-3\]\[q=-2\]This tells us the new inflection point is at \((3,-2)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph A.
Equation 2
The curve’s equation is \(y=-\sqrt[3]{-(x+3)}+2\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(-1)(x+(3))}+(2)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=3\]\[q=2\]This tells us the new inflection point is at \((-3,2)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting both horizontally and vertically results in a return to the original increasing orientation. Thus, the correct graph is graph B.
Equation 3
The curve’s equation is \(y=-\sqrt[3]{x+3}-2\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(1)(x+(3))}+(-2)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=3\]\[q=-2\]This tells us the new inflection point is at \((-3,-2)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph D.
Equation 4
The curve’s equation is \(y=\sqrt[3]{x-3}+2\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(1)(x+(-3))}+(2)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-3\]\[q=2\]This tells us the new inflection point is at \((3,2)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). Because there are no reflections, the curve is in the original increasing orientation. Thus, the correct graph is graph C.
Question
Consider the \(y=\sqrt[3]{x}\) curve (cube-root function) shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 4 graphs are made by translating and reflecting the \(y=\sqrt[3]{x}\) curve. So each one is a different daughter function, made by altering the parent function.
The origin of the parent cube-root function is called the “inflection point”. It is useful to identify where the inflection point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt[3]{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=\sqrt[3]{x-3}+5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(1)(x+(-3))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-3\]\[q=5\]This tells us the new inflection point is at \((3,5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). Because there are no reflections, the curve is in the original increasing orientation. Thus, the correct graph is graph C.
Equation 2
The curve’s equation is \(y=-\sqrt[3]{-(x+3)}-5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(-1)(x+(3))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=3\]\[q=-5\]This tells us the new inflection point is at \((-3,-5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting both horizontally and vertically results in a return to the original increasing orientation. Thus, the correct graph is graph D.
Equation 3
The curve’s equation is \(y=\sqrt[3]{x-3}-5\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(1)(x+(-3))}+(-5)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-3\]\[q=-5\]This tells us the new inflection point is at \((3,-5)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). Because there are no reflections, the curve is in the original increasing orientation. Thus, the correct graph is graph A.
Equation 4
The curve’s equation is \(y=-\sqrt[3]{x+3}+5\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(1)(x+(3))}+(5)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=3\]\[q=5\]This tells us the new inflection point is at \((-3,5)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph B.
Question
Consider the \(y=\sqrt[3]{x}\) curve (cube-root function) shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 4 graphs are made by translating and reflecting the \(y=\sqrt[3]{x}\) curve. So each one is a different daughter function, made by altering the parent function.
Match the 4 graphs with the equations.
Equation \(y=-\sqrt[3]{x+2}+3\) matches graph .
Equation \(y=\sqrt[3]{x-2}-3\) matches graph .
Equation \(y=\sqrt[3]{x-2}+3\) matches graph .
Equation \(y=\sqrt[3]{x+2}-3\) matches graph .
Solution
The origin of the parent cube-root function is called the “inflection point”. It is useful to identify where the inflection point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt[3]{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=-\sqrt[3]{x+2}+3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(1)(x+(2))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=2\]\[q=3\]This tells us the new inflection point is at \((-2,3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph B.
Equation 2
The curve’s equation is \(y=\sqrt[3]{x-2}-3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(1)(x+(-2))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-2\]\[q=-3\]This tells us the new inflection point is at \((2,-3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). Because there are no reflections, the curve is in the original increasing orientation. Thus, the correct graph is graph D.
Equation 3
The curve’s equation is \(y=\sqrt[3]{x-2}+3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(1)(x+(-2))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-2\]\[q=3\]This tells us the new inflection point is at \((2,3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). Because there are no reflections, the curve is in the original increasing orientation. Thus, the correct graph is graph C.
Equation 4
The curve’s equation is \(y=\sqrt[3]{x+2}-3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(1)(x+(2))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=2\]\[q=-3\]This tells us the new inflection point is at \((-2,-3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). Because there are no reflections, the curve is in the original increasing orientation. Thus, the correct graph is graph A.
Question
Consider the \(y=\sqrt[3]{x}\) curve (cube-root function) shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 4 graphs are made by translating and reflecting the \(y=\sqrt[3]{x}\) curve. So each one is a different daughter function, made by altering the parent function.
The origin of the parent cube-root function is called the “inflection point”. It is useful to identify where the inflection point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt[3]{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=\sqrt[3]{-(x+5)}-4\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(-1)(x+(5))}+(-4)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=5\]\[q=-4\]This tells us the new inflection point is at \((-5,-4)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph A.
Equation 2
The curve’s equation is \(y=-\sqrt[3]{-(x-5)}+4\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(-1)(x+(-5))}+(4)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=-5\]\[q=4\]This tells us the new inflection point is at \((5,4)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting both horizontally and vertically results in a return to the original increasing orientation. Thus, the correct graph is graph B.
Equation 3
The curve’s equation is \(y=-\sqrt[3]{x-5}-4\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(1)(x+(-5))}+(-4)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=1\]\[p=-5\]\[q=-4\]This tells us the new inflection point is at \((5,-4)\). We also know there is a vertical reflection (because \(m=-1\)), and there is NOT a horizontal reflection (because \(n=1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph C.
Equation 4
The curve’s equation is \(y=-\sqrt[3]{-(x-5)}-4\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(-1)(x+(-5))}+(-4)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=-5\]\[q=-4\]This tells us the new inflection point is at \((5,-4)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting both horizontally and vertically results in a return to the original increasing orientation. Thus, the correct graph is graph D.
Question
Consider the \(y=\sqrt[3]{x}\) curve (cube-root function) shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 4 graphs are made by translating and reflecting the \(y=\sqrt[3]{x}\) curve. So each one is a different daughter function, made by altering the parent function.
Match the 4 graphs with the equations.
Equation \(y=\sqrt[3]{x-2}+4\) matches graph .
Equation \(y=\sqrt[3]{x+2}-4\) matches graph .
Equation \(y=\sqrt[3]{-(x+2)}-4\) matches graph .
Equation \(y=\sqrt[3]{-(x-2)}-4\) matches graph .
Solution
The origin of the parent cube-root function is called the “inflection point”. It is useful to identify where the inflection point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt[3]{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=\sqrt[3]{x-2}+4\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(1)(x+(-2))}+(4)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=-2\]\[q=4\]This tells us the new inflection point is at \((2,4)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). Because there are no reflections, the curve is in the original increasing orientation. Thus, the correct graph is graph A.
Equation 2
The curve’s equation is \(y=\sqrt[3]{x+2}-4\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(1)(x+(2))}+(-4)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=1\]\[p=2\]\[q=-4\]This tells us the new inflection point is at \((-2,-4)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is NOT a horizontal reflection (because \(n=1\)). Because there are no reflections, the curve is in the original increasing orientation. Thus, the correct graph is graph D.
Equation 3
The curve’s equation is \(y=\sqrt[3]{-(x+2)}-4\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(-1)(x+(2))}+(-4)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=2\]\[q=-4\]This tells us the new inflection point is at \((-2,-4)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph C.
Equation 4
The curve’s equation is \(y=\sqrt[3]{-(x-2)}-4\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(-1)(x+(-2))}+(-4)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=-2\]\[q=-4\]This tells us the new inflection point is at \((2,-4)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph B.
Question
Consider the \(y=\sqrt[3]{x}\) curve (cube-root function) shown below. In the context of transforming functions, a very simple function is often called a “parent” function.
The following 4 graphs are made by translating and reflecting the \(y=\sqrt[3]{x}\) curve. So each one is a different daughter function, made by altering the parent function.
The origin of the parent cube-root function is called the “inflection point”. It is useful to identify where the inflection point gets moved to.
Operations inside the function (under the radical) change the \(x\) values, so they cause horizontal transformations.
Operations outside the function change the \(y\) values, so they cause vertical transformations.
Addition/subtraction causes a translation (shift). Horizontal shifts are counterintuitive.
Multiplying by \(-1\) causes a reflection.
You can account for all the transformations with 4 parameters (\(m\), \(n\), \(p\), and \(q\)).
\[y=m\sqrt[3]{n(x+p)}+q\]
The parameter \(m\) is either \(-1\) or \(1\). If \(m=-1\), then the curve is reflected vertically, so the curve goes down from the starting point (instead of up).
The parameter \(n\) is either \(-1\) or \(1\). If \(n=-1\), then the curve is reflected horizontally, so the curve goes left from the starting point (instead of right).
The parameter \(p\) shifts the curve horizontally.
The parameter \(q\) shifts the curve vertically.
The starting point will be at \((-p,q)\).
Equation 1
The curve’s equation is \(y=-\sqrt[3]{-(x-4)}+3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(-1)(x+(-4))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=-4\]\[q=3\]This tells us the new inflection point is at \((4,3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting both horizontally and vertically results in a return to the original increasing orientation. Thus, the correct graph is graph B.
Equation 2
The curve’s equation is \(y=\sqrt[3]{-(x+4)}-3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(-1)(x+(4))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=4\]\[q=-3\]This tells us the new inflection point is at \((-4,-3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph D.
Equation 3
The curve’s equation is \(y=-\sqrt[3]{-(x+4)}-3\)
Rewrite the equation with explicit parameters.
\(y=(-1)\sqrt[3]{(-1)(x+(4))}+(-3)\)
So, we can more easily see the values of the parameters.
\[m=-1\]\[n=-1\]\[p=4\]\[q=-3\]This tells us the new inflection point is at \((-4,-3)\). We also know there is a vertical reflection (because \(m=-1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting both horizontally and vertically results in a return to the original increasing orientation. Thus, the correct graph is graph A.
Equation 4
The curve’s equation is \(y=\sqrt[3]{-(x+4)}+3\)
Rewrite the equation with explicit parameters.
\(y=(1)\sqrt[3]{(-1)(x+(4))}+(3)\)
So, we can more easily see the values of the parameters.
\[m=1\]\[n=-1\]\[p=4\]\[q=3\]This tells us the new inflection point is at \((-4,3)\). We also know there is NOT a vertical reflection (because \(m=1\)), and there is a horizontal reflection (because \(n=-1\)). For the cube-root function, reflecting either horizontally or vertically results in a decreasing orientation. Thus, the correct graph is graph C.
Expand the right side (by using the FOIL method or the box method).
\[x+7=x^2-10x+25\]
Subtract \(x+7\) from both sides.
\[0=x^2-11x+18\]
Factor the quadratic expression (by guessing and checking).
\[0=(x-2)(x-9)\]
Use the zero product property to find the potential solutions are \(x=2\) and \(x=9\). Plug each potential solution into the original equation to check whether the potential solution is an actual solution or an extraneous solution.
First check whether \(x=2\) satisfies the original equation.
\[\sqrt{(2)+7}+1\stackrel{?}{=}(2)-4\]\[\sqrt{9}+1\stackrel{?}{=}-2\]\[3+1\stackrel{?}{=}-2\]\[4\ne-2\]
So, \(x=2\) is an extraneous solution.
Next, check whether \(x=9\) satisfies the original equation.
\[\sqrt{(9)+7}+1\stackrel{?}{=}(9)-4\]\[\sqrt{16}+1\stackrel{?}{=}5\]\[4+1\stackrel{?}{=}5\]\[5=5\]
So, \(x=9\) is an actual solution.
Graphical interpretation
You can graph \(y=\sqrt{x+7}+1\) and \(y=x-4\) as two curves on a Cartesian-coordinate plane. This will show the actual solution as the horizontal coordinate of the interception. The extraneous solution can be seen as the horizontal coordinate of the intersection of the line with a reflection of the square-root curve that would make a full horizontal parabola (curve \(y=-\sqrt{x+7}+1\)).
Based on the \(x\) values of the intersections, the actual solution is \(x=9\) and the extraneous solution is \(x=2\).
Expand the right side (by using the FOIL method or the box method).
\[x+3=x^2+2x+1\]
Subtract \(x+3\) from both sides.
\[0=x^2+x-2\]
Factor the quadratic expression (by guessing and checking).
\[0=(x-1)(x+2)\]
Use the zero product property to find the potential solutions are \(x=1\) and \(x=-2\). Plug each potential solution into the original equation to check whether the potential solution is an actual solution or an extraneous solution.
First check whether \(x=1\) satisfies the original equation.
\[\sqrt{(1)+3}-3\stackrel{?}{=}-(1)-4\]\[\sqrt{4}-3\stackrel{?}{=}-5\]\[2-3\stackrel{?}{=}-5\]\[-1\ne-5\]
So, \(x=1\) is an extraneous solution.
Next, check whether \(x=-2\) satisfies the original equation.
\[\sqrt{(-2)+3}-3\stackrel{?}{=}-(-2)-4\]\[\sqrt{1}-3\stackrel{?}{=}-2\]\[1-3\stackrel{?}{=}-2\]\[-2=-2\]
So, \(x=-2\) is an actual solution.
Graphical interpretation
You can graph \(y=\sqrt{x+3}-3\) and \(y=-x-4\) as two curves on a Cartesian-coordinate plane. This will show the actual solution as the horizontal coordinate of the interception. The extraneous solution can be seen as the horizontal coordinate of the intersection of the line with a reflection of the square-root curve that would make a full horizontal parabola (curve \(y=-\sqrt{x+3}-3\)).
Based on the \(x\) values of the intersections, the actual solution is \(x=-2\) and the extraneous solution is \(x=1\).
Expand the right side (by using the FOIL method or the box method).
\[x+8=x^2+4x+4\]
Subtract \(x+8\) from both sides.
\[0=x^2+3x-4\]
Factor the quadratic expression (by guessing and checking).
\[0=(x+4)(x-1)\]
Use the zero product property to find the potential solutions are \(x=-4\) and \(x=1\). Plug each potential solution into the original equation to check whether the potential solution is an actual solution or an extraneous solution.
First check whether \(x=-4\) satisfies the original equation.
\[\sqrt{(-4)+8}+5\stackrel{?}{=}-(-4)+3\]\[\sqrt{4}+5\stackrel{?}{=}7\]\[2+5\stackrel{?}{=}7\]\[7=7\]
So, \(x=-4\) is an actual solution.
Next, check whether \(x=1\) satisfies the original equation.
\[\sqrt{(1)+8}+5\stackrel{?}{=}-(1)+3\]\[\sqrt{9}+5\stackrel{?}{=}2\]\[3+5\stackrel{?}{=}2\]\[8\ne2\]
So, \(x=1\) is an extraneous solution.
Graphical interpretation
You can graph \(y=\sqrt{x+8}+5\) and \(y=-x+3\) as two curves on a Cartesian-coordinate plane. This will show the actual solution as the horizontal coordinate of the interception. The extraneous solution can be seen as the horizontal coordinate of the intersection of the line with a reflection of the square-root curve that would make a full horizontal parabola (curve \(y=-\sqrt{x+8}+5\)).
Based on the \(x\) values of the intersections, the actual solution is \(x=-4\) and the extraneous solution is \(x=1\).
Expand the right side (by using the FOIL method or the box method).
\[x+7=x^2+10x+25\]
Subtract \(x+7\) from both sides.
\[0=x^2+9x+18\]
Factor the quadratic expression (by guessing and checking).
\[0=(x+6)(x+3)\]
Use the zero product property to find the potential solutions are \(x=-6\) and \(x=-3\). Plug each potential solution into the original equation to check whether the potential solution is an actual solution or an extraneous solution.
First check whether \(x=-6\) satisfies the original equation.
\[\sqrt{(-6)+7}+6\stackrel{?}{=}-(-6)+1\]\[\sqrt{1}+6\stackrel{?}{=}7\]\[1+6\stackrel{?}{=}7\]\[7=7\]
So, \(x=-6\) is an actual solution.
Next, check whether \(x=-3\) satisfies the original equation.
\[\sqrt{(-3)+7}+6\stackrel{?}{=}-(-3)+1\]\[\sqrt{4}+6\stackrel{?}{=}4\]\[2+6\stackrel{?}{=}4\]\[8\ne4\]
So, \(x=-3\) is an extraneous solution.
Graphical interpretation
You can graph \(y=\sqrt{x+7}+6\) and \(y=-x+1\) as two curves on a Cartesian-coordinate plane. This will show the actual solution as the horizontal coordinate of the interception. The extraneous solution can be seen as the horizontal coordinate of the intersection of the line with a reflection of the square-root curve that would make a full horizontal parabola (curve \(y=-\sqrt{x+7}+6\)).
Based on the \(x\) values of the intersections, the actual solution is \(x=-6\) and the extraneous solution is \(x=-3\).
Expand the right side (by using the FOIL method or the box method).
\[x-3=x^2-10x+25\]
Subtract \(x-3\) from both sides.
\[0=x^2-11x+28\]
Factor the quadratic expression (by guessing and checking).
\[0=(x-7)(x-4)\]
Use the zero product property to find the potential solutions are \(x=7\) and \(x=4\). Plug each potential solution into the original equation to check whether the potential solution is an actual solution or an extraneous solution.
First check whether \(x=7\) satisfies the original equation.
\[\sqrt{(7)-3}+6\stackrel{?}{=}-(7)+11\]\[\sqrt{4}+6\stackrel{?}{=}4\]\[2+6\stackrel{?}{=}4\]\[8\ne4\]
So, \(x=7\) is an extraneous solution.
Next, check whether \(x=4\) satisfies the original equation.
\[\sqrt{(4)-3}+6\stackrel{?}{=}-(4)+11\]\[\sqrt{1}+6\stackrel{?}{=}7\]\[1+6\stackrel{?}{=}7\]\[7=7\]
So, \(x=4\) is an actual solution.
Graphical interpretation
You can graph \(y=\sqrt{x-3}+6\) and \(y=-x+11\) as two curves on a Cartesian-coordinate plane. This will show the actual solution as the horizontal coordinate of the interception. The extraneous solution can be seen as the horizontal coordinate of the intersection of the line with a reflection of the square-root curve that would make a full horizontal parabola (curve \(y=-\sqrt{x-3}+6\)).
Based on the \(x\) values of the intersections, the actual solution is \(x=4\) and the extraneous solution is \(x=7\).
Expand the right side (by using the FOIL method or the box method).
\[x+8=x^2-8x+16\]
Subtract \(x+8\) from both sides.
\[0=x^2-9x+8\]
Factor the quadratic expression (by guessing and checking).
\[0=(x-8)(x-1)\]
Use the zero product property to find the potential solutions are \(x=8\) and \(x=1\). Plug each potential solution into the original equation to check whether the potential solution is an actual solution or an extraneous solution.
First check whether \(x=8\) satisfies the original equation.
\[\sqrt{(8)+8}+1\stackrel{?}{=}(8)-3\]\[\sqrt{16}+1\stackrel{?}{=}5\]\[4+1\stackrel{?}{=}5\]\[5=5\]
So, \(x=8\) is an actual solution.
Next, check whether \(x=1\) satisfies the original equation.
\[\sqrt{(1)+8}+1\stackrel{?}{=}(1)-3\]\[\sqrt{9}+1\stackrel{?}{=}-2\]\[3+1\stackrel{?}{=}-2\]\[4\ne-2\]
So, \(x=1\) is an extraneous solution.
Graphical interpretation
You can graph \(y=\sqrt{x+8}+1\) and \(y=x-3\) as two curves on a Cartesian-coordinate plane. This will show the actual solution as the horizontal coordinate of the interception. The extraneous solution can be seen as the horizontal coordinate of the intersection of the line with a reflection of the square-root curve that would make a full horizontal parabola (curve \(y=-\sqrt{x+8}+1\)).
Based on the \(x\) values of the intersections, the actual solution is \(x=8\) and the extraneous solution is \(x=1\).
Expand the right side (by using the FOIL method or the box method).
\[x-5=x^2-14x+49\]
Subtract \(x-5\) from both sides.
\[0=x^2-15x+54\]
Factor the quadratic expression (by guessing and checking).
\[0=(x-9)(x-6)\]
Use the zero product property to find the potential solutions are \(x=9\) and \(x=6\). Plug each potential solution into the original equation to check whether the potential solution is an actual solution or an extraneous solution.
First check whether \(x=9\) satisfies the original equation.
\[\sqrt{(9)-5}+8\stackrel{?}{=}-(9)+15\]\[\sqrt{4}+8\stackrel{?}{=}6\]\[2+8\stackrel{?}{=}6\]\[10\ne6\]
So, \(x=9\) is an extraneous solution.
Next, check whether \(x=6\) satisfies the original equation.
\[\sqrt{(6)-5}+8\stackrel{?}{=}-(6)+15\]\[\sqrt{1}+8\stackrel{?}{=}9\]\[1+8\stackrel{?}{=}9\]\[9=9\]
So, \(x=6\) is an actual solution.
Graphical interpretation
You can graph \(y=\sqrt{x-5}+8\) and \(y=-x+15\) as two curves on a Cartesian-coordinate plane. This will show the actual solution as the horizontal coordinate of the interception. The extraneous solution can be seen as the horizontal coordinate of the intersection of the line with a reflection of the square-root curve that would make a full horizontal parabola (curve \(y=-\sqrt{x-5}+8\)).
Based on the \(x\) values of the intersections, the actual solution is \(x=6\) and the extraneous solution is \(x=9\).
Expand the right side (by using the FOIL method or the box method).
\[x+7=x^2+10x+25\]
Subtract \(x+7\) from both sides.
\[0=x^2+9x+18\]
Factor the quadratic expression (by guessing and checking).
\[0=(x+3)(x+6)\]
Use the zero product property to find the potential solutions are \(x=-3\) and \(x=-6\). Plug each potential solution into the original equation to check whether the potential solution is an actual solution or an extraneous solution.
First check whether \(x=-3\) satisfies the original equation.
\[\sqrt{(-3)+7}+2\stackrel{?}{=}(-3)+7\]\[\sqrt{4}+2\stackrel{?}{=}4\]\[2+2\stackrel{?}{=}4\]\[4=4\]
So, \(x=-3\) is an actual solution.
Next, check whether \(x=-6\) satisfies the original equation.
\[\sqrt{(-6)+7}+2\stackrel{?}{=}(-6)+7\]\[\sqrt{1}+2\stackrel{?}{=}1\]\[1+2\stackrel{?}{=}1\]\[3\ne1\]
So, \(x=-6\) is an extraneous solution.
Graphical interpretation
You can graph \(y=\sqrt{x+7}+2\) and \(y=x+7\) as two curves on a Cartesian-coordinate plane. This will show the actual solution as the horizontal coordinate of the interception. The extraneous solution can be seen as the horizontal coordinate of the intersection of the line with a reflection of the square-root curve that would make a full horizontal parabola (curve \(y=-\sqrt{x+7}+2\)).
Based on the \(x\) values of the intersections, the actual solution is \(x=-3\) and the extraneous solution is \(x=-6\).
Expand the right side (by using the FOIL method or the box method).
\[x-2=x^2-8x+16\]
Subtract \(x-2\) from both sides.
\[0=x^2-9x+18\]
Factor the quadratic expression (by guessing and checking).
\[0=(x-6)(x-3)\]
Use the zero product property to find the potential solutions are \(x=6\) and \(x=3\). Plug each potential solution into the original equation to check whether the potential solution is an actual solution or an extraneous solution.
First check whether \(x=6\) satisfies the original equation.
\[\sqrt{(6)-2}+4\stackrel{?}{=}-(6)+8\]\[\sqrt{4}+4\stackrel{?}{=}2\]\[2+4\stackrel{?}{=}2\]\[6\ne2\]
So, \(x=6\) is an extraneous solution.
Next, check whether \(x=3\) satisfies the original equation.
\[\sqrt{(3)-2}+4\stackrel{?}{=}-(3)+8\]\[\sqrt{1}+4\stackrel{?}{=}5\]\[1+4\stackrel{?}{=}5\]\[5=5\]
So, \(x=3\) is an actual solution.
Graphical interpretation
You can graph \(y=\sqrt{x-2}+4\) and \(y=-x+8\) as two curves on a Cartesian-coordinate plane. This will show the actual solution as the horizontal coordinate of the interception. The extraneous solution can be seen as the horizontal coordinate of the intersection of the line with a reflection of the square-root curve that would make a full horizontal parabola (curve \(y=-\sqrt{x-2}+4\)).
Based on the \(x\) values of the intersections, the actual solution is \(x=3\) and the extraneous solution is \(x=6\).
Expand the right side (by using the FOIL method or the box method).
\[x-1=x^2-6x+9\]
Subtract \(x-1\) from both sides.
\[0=x^2-7x+10\]
Factor the quadratic expression (by guessing and checking).
\[0=(x-2)(x-5)\]
Use the zero product property to find the potential solutions are \(x=2\) and \(x=5\). Plug each potential solution into the original equation to check whether the potential solution is an actual solution or an extraneous solution.
First check whether \(x=2\) satisfies the original equation.
\[\sqrt{(2)-1}-8\stackrel{?}{=}(2)-11\]\[\sqrt{1}-8\stackrel{?}{=}-9\]\[1-8\stackrel{?}{=}-9\]\[-7\ne-9\]
So, \(x=2\) is an extraneous solution.
Next, check whether \(x=5\) satisfies the original equation.
\[\sqrt{(5)-1}-8\stackrel{?}{=}(5)-11\]\[\sqrt{4}-8\stackrel{?}{=}-6\]\[2-8\stackrel{?}{=}-6\]\[-6=-6\]
So, \(x=5\) is an actual solution.
Graphical interpretation
You can graph \(y=\sqrt{x-1}-8\) and \(y=x-11\) as two curves on a Cartesian-coordinate plane. This will show the actual solution as the horizontal coordinate of the interception. The extraneous solution can be seen as the horizontal coordinate of the intersection of the line with a reflection of the square-root curve that would make a full horizontal parabola (curve \(y=-\sqrt{x-1}-8\)).
Based on the \(x\) values of the intersections, the actual solution is \(x=5\) and the extraneous solution is \(x=2\).
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{(x-2)(x-5)}{(x-1)(x-5)}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
The \(x\)-intercept is at \(x=2\).
\[f(2)~=~\frac{(2-2)(2-5)}{(2-1)(2-5)} ~=~ \frac{(0)(-3)}{(1)(-3)} ~=~ \frac{0}{-3}\]
The hole is at \(x=5\).
\[f(5)~=~\frac{(5-2)(5-5)}{(5-1)(5-5)} ~=~ \frac{(3)(0)}{(4)(0)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=1\).
\[f(1)~=~\frac{(1-2)(1-5)}{(1-1)(1-5)} ~=~ \frac{(-1)(-4)}{(0)(-4)} ~=~ \frac{4}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{(x+6)(x-1)}{(x+6)(x-2)}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
The \(x\)-intercept is at \(x=1\).
\[f(1)~=~\frac{(1+6)(1-1)}{(1+6)(1-2)} ~=~ \frac{(7)(0)}{(7)(-1)} ~=~ \frac{0}{-7}\]
The hole is at \(x=-6\).
\[f(-6)~=~\frac{(-6+6)(-6-1)}{(-6+6)(-6-2)} ~=~ \frac{(0)(-7)}{(0)(-8)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=2\).
\[f(2)~=~\frac{(2+6)(2-1)}{(2+6)(2-2)} ~=~ \frac{(8)(1)}{(8)(0)} ~=~ \frac{8}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{(x+1)(x-3)}{(x+1)(x-4)}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
The \(x\)-intercept is at \(x=3\).
\[f(3)~=~\frac{(3+1)(3-3)}{(3+1)(3-4)} ~=~ \frac{(4)(0)}{(4)(-1)} ~=~ \frac{0}{-4}\]
The hole is at \(x=-1\).
\[f(-1)~=~\frac{(-1+1)(-1-3)}{(-1+1)(-1-4)} ~=~ \frac{(0)(-4)}{(0)(-5)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=4\).
\[f(4)~=~\frac{(4+1)(4-3)}{(4+1)(4-4)} ~=~ \frac{(5)(1)}{(5)(0)} ~=~ \frac{5}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{(x-2)(x-3)}{(x+5)(x-2)}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
The \(x\)-intercept is at \(x=3\).
\[f(3)~=~\frac{(3-2)(3-3)}{(3+5)(3-2)} ~=~ \frac{(1)(0)}{(8)(1)} ~=~ \frac{0}{8}\]
The hole is at \(x=2\).
\[f(2)~=~\frac{(2-2)(2-3)}{(2+5)(2-2)} ~=~ \frac{(0)(-1)}{(7)(0)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=-5\).
\[f(-5)~=~\frac{(-5-2)(-5-3)}{(-5+5)(-5-2)} ~=~ \frac{(-7)(-8)}{(0)(-7)} ~=~ \frac{56}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{(x+4)(x+3)}{(x+3)(x-2)}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
The \(x\)-intercept is at \(x=-4\).
\[f(-4)~=~\frac{(-4+4)(-4+3)}{(-4+3)(-4-2)} ~=~ \frac{(0)(-1)}{(-1)(-6)} ~=~ \frac{0}{6}\]
The hole is at \(x=-3\).
\[f(-3)~=~\frac{(-3+4)(-3+3)}{(-3+3)(-3-2)} ~=~ \frac{(1)(0)}{(0)(-5)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=2\).
\[f(2)~=~\frac{(2+4)(2+3)}{(2+3)(2-2)} ~=~ \frac{(6)(5)}{(5)(0)} ~=~ \frac{30}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{(x+6)(x+4)}{(x+6)(x+5)}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
The \(x\)-intercept is at \(x=-4\).
\[f(-4)~=~\frac{(-4+6)(-4+4)}{(-4+6)(-4+5)} ~=~ \frac{(2)(0)}{(2)(1)} ~=~ \frac{0}{2}\]
The hole is at \(x=-6\).
\[f(-6)~=~\frac{(-6+6)(-6+4)}{(-6+6)(-6+5)} ~=~ \frac{(0)(-2)}{(0)(-1)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=-5\).
\[f(-5)~=~\frac{(-5+6)(-5+4)}{(-5+6)(-5+5)} ~=~ \frac{(1)(-1)}{(1)(0)} ~=~ \frac{-1}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{(x+3)(x-2)}{(x+3)(x-6)}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
The \(x\)-intercept is at \(x=2\).
\[f(2)~=~\frac{(2+3)(2-2)}{(2+3)(2-6)} ~=~ \frac{(5)(0)}{(5)(-4)} ~=~ \frac{0}{-20}\]
The hole is at \(x=-3\).
\[f(-3)~=~\frac{(-3+3)(-3-2)}{(-3+3)(-3-6)} ~=~ \frac{(0)(-5)}{(0)(-9)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=6\).
\[f(6)~=~\frac{(6+3)(6-2)}{(6+3)(6-6)} ~=~ \frac{(9)(4)}{(9)(0)} ~=~ \frac{36}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{(x+6)(x-5)}{(x-1)(x-5)}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
The \(x\)-intercept is at \(x=-6\).
\[f(-6)~=~\frac{(-6+6)(-6-5)}{(-6-1)(-6-5)} ~=~ \frac{(0)(-11)}{(-7)(-11)} ~=~ \frac{0}{77}\]
The hole is at \(x=5\).
\[f(5)~=~\frac{(5+6)(5-5)}{(5-1)(5-5)} ~=~ \frac{(11)(0)}{(4)(0)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=1\).
\[f(1)~=~\frac{(1+6)(1-5)}{(1-1)(1-5)} ~=~ \frac{(7)(-4)}{(0)(-4)} ~=~ \frac{-28}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{(x+2)(x-3)}{(x+1)(x-3)}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
The \(x\)-intercept is at \(x=-2\).
\[f(-2)~=~\frac{(-2+2)(-2-3)}{(-2+1)(-2-3)} ~=~ \frac{(0)(-5)}{(-1)(-5)} ~=~ \frac{0}{5}\]
The hole is at \(x=3\).
\[f(3)~=~\frac{(3+2)(3-3)}{(3+1)(3-3)} ~=~ \frac{(5)(0)}{(4)(0)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=-1\).
\[f(-1)~=~\frac{(-1+2)(-1-3)}{(-1+1)(-1-3)} ~=~ \frac{(1)(-4)}{(0)(-4)} ~=~ \frac{-4}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{(x+6)(x-1)}{(x+5)(x-1)}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
The \(x\)-intercept is at \(x=-6\).
\[f(-6)~=~\frac{(-6+6)(-6-1)}{(-6+5)(-6-1)} ~=~ \frac{(0)(-7)}{(-1)(-7)} ~=~ \frac{0}{7}\]
The hole is at \(x=1\).
\[f(1)~=~\frac{(1+6)(1-1)}{(1+5)(1-1)} ~=~ \frac{(7)(0)}{(6)(0)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=-5\).
\[f(-5)~=~\frac{(-5+6)(-5-1)}{(-5+5)(-5-1)} ~=~ \frac{(1)(-6)}{(0)(-6)} ~=~ \frac{-6}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
Start by factoring the quadratic expressions in the numerator and the denominator.
\[f(x)~=~\frac{(x-3)(x-6)}{(x-2)(x-6)}\]
The \(x\)-intercept is at \(x=3\).
\[f(3)~=~\frac{(3-3)(3-6)}{(3-2)(3-6)} ~=~ \frac{(0)(-3)}{(1)(-3)} ~=~ \frac{0}{-3}\]
The hole is at \(x=6\).
\[f(6)~=~\frac{(6-3)(6-6)}{(6-2)(6-6)} ~=~ \frac{(3)(0)}{(4)(0)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=2\).
\[f(2)~=~\frac{(2-3)(2-6)}{(2-2)(2-6)} ~=~ \frac{(-1)(-4)}{(0)(-4)} ~=~ \frac{4}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{x^{2}-{2}x-{24}}{x^{2}-x-{30}}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
Start by factoring the quadratic expressions in the numerator and the denominator.
\[f(x)~=~\frac{(x+4)(x-6)}{(x+5)(x-6)}\]
The \(x\)-intercept is at \(x=-4\).
\[f(-4)~=~\frac{(-4+4)(-4-6)}{(-4+5)(-4-6)} ~=~ \frac{(0)(-10)}{(1)(-10)} ~=~ \frac{0}{-10}\]
The hole is at \(x=6\).
\[f(6)~=~\frac{(6+4)(6-6)}{(6+5)(6-6)} ~=~ \frac{(10)(0)}{(11)(0)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=-5\).
\[f(-5)~=~\frac{(-5+4)(-5-6)}{(-5+5)(-5-6)} ~=~ \frac{(-1)(-11)}{(0)(-11)} ~=~ \frac{11}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{x^{2}-{5}x+{4}}{x^{2}-{6}x+{8}}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
Start by factoring the quadratic expressions in the numerator and the denominator.
\[f(x)~=~\frac{(x-1)(x-4)}{(x-2)(x-4)}\]
The \(x\)-intercept is at \(x=1\).
\[f(1)~=~\frac{(1-1)(1-4)}{(1-2)(1-4)} ~=~ \frac{(0)(-3)}{(-1)(-3)} ~=~ \frac{0}{3}\]
The hole is at \(x=4\).
\[f(4)~=~\frac{(4-1)(4-4)}{(4-2)(4-4)} ~=~ \frac{(3)(0)}{(2)(0)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=2\).
\[f(2)~=~\frac{(2-1)(2-4)}{(2-2)(2-4)} ~=~ \frac{(1)(-2)}{(0)(-2)} ~=~ \frac{-2}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
Start by factoring the quadratic expressions in the numerator and the denominator.
\[f(x)~=~\frac{(x+4)(x+2)}{(x+6)(x+4)}\]
The \(x\)-intercept is at \(x=-2\).
\[f(-2)~=~\frac{(-2+4)(-2+2)}{(-2+6)(-2+4)} ~=~ \frac{(2)(0)}{(4)(2)} ~=~ \frac{0}{8}\]
The hole is at \(x=-4\).
\[f(-4)~=~\frac{(-4+4)(-4+2)}{(-4+6)(-4+4)} ~=~ \frac{(0)(-2)}{(2)(0)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=-6\).
\[f(-6)~=~\frac{(-6+4)(-6+2)}{(-6+6)(-6+4)} ~=~ \frac{(-2)(-4)}{(0)(-2)} ~=~ \frac{8}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{x^{2}+{2}x-{15}}{x^{2}+{6}x+{5}}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
Start by factoring the quadratic expressions in the numerator and the denominator.
\[f(x)~=~\frac{(x+5)(x-3)}{(x+5)(x+1)}\]
The \(x\)-intercept is at \(x=3\).
\[f(3)~=~\frac{(3+5)(3-3)}{(3+5)(3+1)} ~=~ \frac{(8)(0)}{(8)(4)} ~=~ \frac{0}{32}\]
The hole is at \(x=-5\).
\[f(-5)~=~\frac{(-5+5)(-5-3)}{(-5+5)(-5+1)} ~=~ \frac{(0)(-8)}{(0)(-4)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=-1\).
\[f(-1)~=~\frac{(-1+5)(-1-3)}{(-1+5)(-1+1)} ~=~ \frac{(4)(-4)}{(4)(0)} ~=~ \frac{-16}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{x^{2}+{2}x-{24}}{x^{2}+x-{30}}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
Start by factoring the quadratic expressions in the numerator and the denominator.
\[f(x)~=~\frac{(x+6)(x-4)}{(x+6)(x-5)}\]
The \(x\)-intercept is at \(x=4\).
\[f(4)~=~\frac{(4+6)(4-4)}{(4+6)(4-5)} ~=~ \frac{(10)(0)}{(10)(-1)} ~=~ \frac{0}{-10}\]
The hole is at \(x=-6\).
\[f(-6)~=~\frac{(-6+6)(-6-4)}{(-6+6)(-6-5)} ~=~ \frac{(0)(-10)}{(0)(-11)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=5\).
\[f(5)~=~\frac{(5+6)(5-4)}{(5+6)(5-5)} ~=~ \frac{(11)(1)}{(11)(0)} ~=~ \frac{11}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{x^{2}-{6}x+{5}}{x^{2}-{7}x+{10}}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
Start by factoring the quadratic expressions in the numerator and the denominator.
\[f(x)~=~\frac{(x-1)(x-5)}{(x-2)(x-5)}\]
The \(x\)-intercept is at \(x=1\).
\[f(1)~=~\frac{(1-1)(1-5)}{(1-2)(1-5)} ~=~ \frac{(0)(-4)}{(-1)(-4)} ~=~ \frac{0}{4}\]
The hole is at \(x=5\).
\[f(5)~=~\frac{(5-1)(5-5)}{(5-2)(5-5)} ~=~ \frac{(4)(0)}{(3)(0)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=2\).
\[f(2)~=~\frac{(2-1)(2-5)}{(2-2)(2-5)} ~=~ \frac{(1)(-3)}{(0)(-3)} ~=~ \frac{-3}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{x^{2}-{3}x-{10}}{x^{2}+x-{30}}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
Start by factoring the quadratic expressions in the numerator and the denominator.
\[f(x)~=~\frac{(x+2)(x-5)}{(x+6)(x-5)}\]
The \(x\)-intercept is at \(x=-2\).
\[f(-2)~=~\frac{(-2+2)(-2-5)}{(-2+6)(-2-5)} ~=~ \frac{(0)(-7)}{(4)(-7)} ~=~ \frac{0}{-28}\]
The hole is at \(x=5\).
\[f(5)~=~\frac{(5+2)(5-5)}{(5+6)(5-5)} ~=~ \frac{(7)(0)}{(11)(0)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=-6\).
\[f(-6)~=~\frac{(-6+2)(-6-5)}{(-6+6)(-6-5)} ~=~ \frac{(-4)(-11)}{(0)(-11)} ~=~ \frac{44}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
\[f(x)~=~\frac{x^{2}+{5}x+{4}}{x^{2}-{2}x-{3}}\]
The \(x\)-intercept is at
The hole is at
The vertical asymptote is at
Solution
Start by factoring the quadratic expressions in the numerator and the denominator.
\[f(x)~=~\frac{(x+4)(x+1)}{(x+1)(x-3)}\]
The \(x\)-intercept is at \(x=-4\).
\[f(-4)~=~\frac{(-4+4)(-4+1)}{(-4+1)(-4-3)} ~=~ \frac{(0)(-3)}{(-3)(-7)} ~=~ \frac{0}{21}\]
The hole is at \(x=-1\).
\[f(-1)~=~\frac{(-1+4)(-1+1)}{(-1+1)(-1-3)} ~=~ \frac{(3)(0)}{(0)(-4)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=3\).
\[f(3)~=~\frac{(3+4)(3+1)}{(3+1)(3-3)} ~=~ \frac{(7)(4)}{(4)(0)} ~=~ \frac{28}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
To identify the \(x\)-intercepts, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }\ne 0\).
To identify the \(x\) values of the holes, find values of \(x\) such that \(\mathrm{numerator} = 0\) and \(\mathrm{denominator }= 0\).
To identify the \(x\) values of the vertical asymptotes, find values of \(x\) such that \(\mathrm{numerator} \ne 0\) and \(\mathrm{denominator }= 0\).
Technically, if factors have multiplicity, it is possible for these simple rules to be “wrong”. Specifically, if the same factor is in both numerator and denominator, but with higher multiplicity in the denominator, the result is a vertical asymptote instead of a hole. This technicality is outside the scope of this class.
The rational function below has an \(x\)-intercept, a hole, and a vertical asymptote. Identify the \(x\) coordinate of each feature.
Start by factoring the quadratic expressions in the numerator and the denominator.
\[f(x)~=~\frac{(x+5)(x+4)}{(x+5)(x-3)}\]
The \(x\)-intercept is at \(x=-4\).
\[f(-4)~=~\frac{(-4+5)(-4+4)}{(-4+5)(-4-3)} ~=~ \frac{(1)(0)}{(1)(-7)} ~=~ \frac{0}{-7}\]
The hole is at \(x=-5\).
\[f(-5)~=~\frac{(-5+5)(-5+4)}{(-5+5)(-5-3)} ~=~ \frac{(0)(-1)}{(0)(-8)} ~=~ \frac{0}{0}\]
The vertical asymptote is at \(x=3\).
\[f(3)~=~\frac{(3+5)(3+4)}{(3+5)(3-3)} ~=~ \frac{(8)(7)}{(8)(0)} ~=~ \frac{56}{0}\]
I’ll point out the hole seems like it could easily be “fixed” by continuing the curve through that point. Indeed, this is of interest. Notice, it is quite easy to write function \(g\) to accomplish this goal by eliminating the common factor in the numerator and denominator. (To be clear, this is not directly helpful for answering the given question.)
Find the \((x,y)\)-coordinates of the hole (removable discontinuity) on the curve \(y=f(x)\).
The hole is at:
(, )
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x-1)(x-8)}{(x-7)(x-8)}\]
Notice, the factor \((x-8)\) is in both numerator and denominator. So, the hole is at \(x=8\). Create a new function \(g\) with a filled in hole by eliminating common factors.
\[g(x)=\frac{x-1}{x-7}\]
Plug in \(x=8\) into function \(g\).
\[g(8)~=~\frac{8-1}{8-7}~=~\frac{7}{1}~=~7\]
Thus, the hole of \(y=f(x)\) is at point \((8,7)\).
Question
Let rational function \(f\) be defined below.
\[f(x)~=~\frac{x^{2}-{8}x+{12}}{x^{2}-{5}x+{6}}\]
Find the \((x,y)\)-coordinates of the hole (removable discontinuity) on the curve \(y=f(x)\).
The hole is at:
(, )
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x-2)(x-6)}{(x-2)(x-3)}\]
Notice, the factor \((x-2)\) is in both numerator and denominator. So, the hole is at \(x=2\). Create a new function \(g\) with a filled in hole by eliminating common factors.
\[g(x)=\frac{x-6}{x-3}\]
Plug in \(x=2\) into function \(g\).
\[g(2)~=~\frac{2-6}{2-3}~=~\frac{-4}{-1}~=~4\]
Thus, the hole of \(y=f(x)\) is at point \((2,4)\).
Find the \((x,y)\)-coordinates of the hole (removable discontinuity) on the curve \(y=f(x)\).
The hole is at:
(, )
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x-2)(x-7)}{(x-7)(x-8)}\]
Notice, the factor \((x-7)\) is in both numerator and denominator. So, the hole is at \(x=7\). Create a new function \(g\) with a filled in hole by eliminating common factors.
\[g(x)=\frac{x-2}{x-8}\]
Plug in \(x=7\) into function \(g\).
\[g(7)~=~\frac{7-2}{7-8}~=~\frac{5}{-1}~=~-5\]
Thus, the hole of \(y=f(x)\) is at point \((7,-5)\).
Find the \((x,y)\)-coordinates of the hole (removable discontinuity) on the curve \(y=f(x)\).
The hole is at:
(, )
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x+4)(x-6)}{(x+9)(x+4)}\]
Notice, the factor \((x+4)\) is in both numerator and denominator. So, the hole is at \(x=-4\). Create a new function \(g\) with a filled in hole by eliminating common factors.
Thus, the hole of \(y=f(x)\) is at point \((-4,-2)\).
Question
Let rational function \(f\) be defined below.
\[f(x)~=~\frac{x^{2}-{6}x+{8}}{x^{2}-{5}x+{6}}\]
Find the \((x,y)\)-coordinates of the hole (removable discontinuity) on the curve \(y=f(x)\).
The hole is at:
(, )
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x-2)(x-4)}{(x-2)(x-3)}\]
Notice, the factor \((x-2)\) is in both numerator and denominator. So, the hole is at \(x=2\). Create a new function \(g\) with a filled in hole by eliminating common factors.
\[g(x)=\frac{x-4}{x-3}\]
Plug in \(x=2\) into function \(g\).
\[g(2)~=~\frac{2-4}{2-3}~=~\frac{-2}{-1}~=~2\]
Thus, the hole of \(y=f(x)\) is at point \((2,2)\).
Find the \((x,y)\)-coordinates of the hole (removable discontinuity) on the curve \(y=f(x)\).
The hole is at:
(, )
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x+10)(x-2)}{(x+10)(x+8)}\]
Notice, the factor \((x+10)\) is in both numerator and denominator. So, the hole is at \(x=-10\). Create a new function \(g\) with a filled in hole by eliminating common factors.
Find the \((x,y)\)-coordinates of the hole (removable discontinuity) on the curve \(y=f(x)\).
The hole is at:
(, )
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x+8)(x-6)}{(x+10)(x+8)}\]
Notice, the factor \((x+8)\) is in both numerator and denominator. So, the hole is at \(x=-8\). Create a new function \(g\) with a filled in hole by eliminating common factors.
Find the \((x,y)\)-coordinates of the hole (removable discontinuity) on the curve \(y=f(x)\).
The hole is at:
(, )
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x-6)(x-7)}{(x-7)(x-8)}\]
Notice, the factor \((x-7)\) is in both numerator and denominator. So, the hole is at \(x=7\). Create a new function \(g\) with a filled in hole by eliminating common factors.
\[g(x)=\frac{x-6}{x-8}\]
Plug in \(x=7\) into function \(g\).
\[g(7)~=~\frac{7-6}{7-8}~=~\frac{1}{-1}~=~-1\]
Thus, the hole of \(y=f(x)\) is at point \((7,-1)\).
Question
Let rational function \(f\) be defined below.
\[f(x)~=~\frac{x^{2}+{9}x+{14}}{x^{2}+{5}x+{6}}\]
Find the \((x,y)\)-coordinates of the hole (removable discontinuity) on the curve \(y=f(x)\).
The hole is at:
(, )
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x+7)(x+2)}{(x+3)(x+2)}\]
Notice, the factor \((x+2)\) is in both numerator and denominator. So, the hole is at \(x=-2\). Create a new function \(g\) with a filled in hole by eliminating common factors.
\[g(x)=\frac{x+7}{x+3}\]
Plug in \(x=-2\) into function \(g\).
\[g(-2)~=~\frac{-2+7}{-2+3}~=~\frac{5}{1}~=~5\]
Thus, the hole of \(y=f(x)\) is at point \((-2,5)\).
Find the \((x,y)\)-coordinates of the hole (removable discontinuity) on the curve \(y=f(x)\).
The hole is at:
(, )
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x-2)(x-10)}{(x-1)(x-2)}\]
Notice, the factor \((x-2)\) is in both numerator and denominator. So, the hole is at \(x=2\). Create a new function \(g\) with a filled in hole by eliminating common factors.
\[g(x)=\frac{x-10}{x-1}\]
Plug in \(x=2\) into function \(g\).
\[g(2)~=~\frac{2-10}{2-1}~=~\frac{-8}{1}~=~-8\]
Thus, the hole of \(y=f(x)\) is at point \((2,-8)\).
Let linear function \(L\) have a slope of 1 and an unknown parameter (\(b\)) dictating the \(y\)-intercept.
\[L(x)~=~x+B\]
Both curves, \(y=f(x)\) and \(y=L(x)\), are graphed. The hole of function \(f\) is at point \((h,k)\), and the point \((h,k)\) lies on the line, so \(k=L(h)\). This implies that the equation \(f(x)=L(x)\) would have an extraneous solution of \(x=h\).
Find the unknown parameter \(B\).
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x+3)(x-7)}{(x-7)(x-8)}\]
Notice, the factor \((x-7)\) is in both numerator and denominator. So, the hole is at \(x=7\). Create a new function \(g\) with a filled in hole by eliminating common factors.
\[g(x)=\frac{x+3}{x-8}\]
Plug in \(x=7\) into function \(g\).
\[g(7)~=~\frac{7+3}{7-8}~=~\frac{10}{-1}~=~-10\]
Thus, the hole of \(y=f(x)\) is at point \((7,-10)\). This tells us that \(-10=L(7)\).
\[-10={7}+{B}\]
Subtract 7 from both sides.
\[-17=B\]\[B=-17\]
Notices this makes the line go through the hole.
You might find it interesting to use algebraic manipulations to solve \(f(x)=L(x)\), in order to see that you’d be led to an extraneous solution.
Which provides two solutions, \(x=7\) and \(x=19\). However, we know that \(x=7\) cannot be a solution, because \(f(7)\) is undefined! Thus, we see that if \(L(x)=x-{17}\), then we get an extraneous solution to \(f(x)=L(x)\).
This reinforces the implications of having \(B=-17\), which makes the line go through the hole.
Question
Let rational function \(f\) be defined below.
\[f(x)~=~\frac{x^{2}+{2}x-{35}}{x^{2}-{6}x+{5}}\]
Let linear function \(L\) have a slope of 1 and an unknown parameter (\(b\)) dictating the \(y\)-intercept.
\[L(x)~=~x+B\]
Both curves, \(y=f(x)\) and \(y=L(x)\), are graphed. The hole of function \(f\) is at point \((h,k)\), and the point \((h,k)\) lies on the line, so \(k=L(h)\). This implies that the equation \(f(x)=L(x)\) would have an extraneous solution of \(x=h\).
Find the unknown parameter \(B\).
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x+7)(x-5)}{(x-1)(x-5)}\]
Notice, the factor \((x-5)\) is in both numerator and denominator. So, the hole is at \(x=5\). Create a new function \(g\) with a filled in hole by eliminating common factors.
\[g(x)=\frac{x+7}{x-1}\]
Plug in \(x=5\) into function \(g\).
\[g(5)~=~\frac{5+7}{5-1}~=~\frac{12}{4}~=~3\]
Thus, the hole of \(y=f(x)\) is at point \((5,3)\). This tells us that \(3=L(5)\).
\[3={5}+{B}\]
Subtract 5 from both sides.
\[-2=B\]\[B=-2\]
Notices this makes the line go through the hole.
You might find it interesting to use algebraic manipulations to solve \(f(x)=L(x)\), in order to see that you’d be led to an extraneous solution.
Which provides two solutions, \(x=-1\) and \(x=5\). However, we know that \(x=5\) cannot be a solution, because \(f(5)\) is undefined! Thus, we see that if \(L(x)=x-{2}\), then we get an extraneous solution to \(f(x)=L(x)\).
This reinforces the implications of having \(B=-2\), which makes the line go through the hole.
Let linear function \(L\) have a slope of 1 and an unknown parameter (\(b\)) dictating the \(y\)-intercept.
\[L(x)~=~x+B\]
Both curves, \(y=f(x)\) and \(y=L(x)\), are graphed. The hole of function \(f\) is at point \((h,k)\), and the point \((h,k)\) lies on the line, so \(k=L(h)\). This implies that the equation \(f(x)=L(x)\) would have an extraneous solution of \(x=h\).
Find the unknown parameter \(B\).
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x-1)(x-8)}{(x-8)(x-9)}\]
Notice, the factor \((x-8)\) is in both numerator and denominator. So, the hole is at \(x=8\). Create a new function \(g\) with a filled in hole by eliminating common factors.
\[g(x)=\frac{x-1}{x-9}\]
Plug in \(x=8\) into function \(g\).
\[g(8)~=~\frac{8-1}{8-9}~=~\frac{7}{-1}~=~-7\]
Thus, the hole of \(y=f(x)\) is at point \((8,-7)\). This tells us that \(-7=L(8)\).
\[-7={8}+{B}\]
Subtract 8 from both sides.
\[-15=B\]\[B=-15\]
Notices this makes the line go through the hole.
You might find it interesting to use algebraic manipulations to solve \(f(x)=L(x)\), in order to see that you’d be led to an extraneous solution.
Which provides two solutions, \(x=8\) and \(x=17\). However, we know that \(x=8\) cannot be a solution, because \(f(8)\) is undefined! Thus, we see that if \(L(x)=x-{15}\), then we get an extraneous solution to \(f(x)=L(x)\).
This reinforces the implications of having \(B=-15\), which makes the line go through the hole.
Let linear function \(L\) have a slope of 1 and an unknown parameter (\(b\)) dictating the \(y\)-intercept.
\[L(x)~=~x+B\]
Both curves, \(y=f(x)\) and \(y=L(x)\), are graphed. The hole of function \(f\) is at point \((h,k)\), and the point \((h,k)\) lies on the line, so \(k=L(h)\). This implies that the equation \(f(x)=L(x)\) would have an extraneous solution of \(x=h\).
Find the unknown parameter \(B\).
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x+5)(x-3)}{(x-3)(x-7)}\]
Notice, the factor \((x-3)\) is in both numerator and denominator. So, the hole is at \(x=3\). Create a new function \(g\) with a filled in hole by eliminating common factors.
\[g(x)=\frac{x+5}{x-7}\]
Plug in \(x=3\) into function \(g\).
\[g(3)~=~\frac{3+5}{3-7}~=~\frac{8}{-4}~=~-2\]
Thus, the hole of \(y=f(x)\) is at point \((3,-2)\). This tells us that \(-2=L(3)\).
\[-2={3}+{B}\]
Subtract 3 from both sides.
\[-5=B\]\[B=-5\]
Notices this makes the line go through the hole.
You might find it interesting to use algebraic manipulations to solve \(f(x)=L(x)\), in order to see that you’d be led to an extraneous solution.
Which provides two solutions, \(x=3\) and \(x=10\). However, we know that \(x=3\) cannot be a solution, because \(f(3)\) is undefined! Thus, we see that if \(L(x)=x-{5}\), then we get an extraneous solution to \(f(x)=L(x)\).
This reinforces the implications of having \(B=-5\), which makes the line go through the hole.
Let linear function \(L\) have a slope of 1 and an unknown parameter (\(b\)) dictating the \(y\)-intercept.
\[L(x)~=~x+B\]
Both curves, \(y=f(x)\) and \(y=L(x)\), are graphed. The hole of function \(f\) is at point \((h,k)\), and the point \((h,k)\) lies on the line, so \(k=L(h)\). This implies that the equation \(f(x)=L(x)\) would have an extraneous solution of \(x=h\).
Find the unknown parameter \(B\).
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x+10)(x+8)}{(x+10)(x+9)}\]
Notice, the factor \((x+10)\) is in both numerator and denominator. So, the hole is at \(x=-10\). Create a new function \(g\) with a filled in hole by eliminating common factors.
Which provides two solutions, \(x=-10\) and \(x=-10\). However, we know that \(x=-10\) cannot be a solution, because \(f(-10)\) is undefined! Thus, we see that if \(L(x)=x+{12}\), then we get an extraneous solution to \(f(x)=L(x)\).
This reinforces the implications of having \(B=12\), which makes the line go through the hole.
Question
Let rational function \(f\) be defined below.
\[f(x)~=~\frac{x^{2}+{6}x+{5}}{x^{2}+{9}x+{20}}\]
Let linear function \(L\) have a slope of 1 and an unknown parameter (\(b\)) dictating the \(y\)-intercept.
\[L(x)~=~x+B\]
Both curves, \(y=f(x)\) and \(y=L(x)\), are graphed. The hole of function \(f\) is at point \((h,k)\), and the point \((h,k)\) lies on the line, so \(k=L(h)\). This implies that the equation \(f(x)=L(x)\) would have an extraneous solution of \(x=h\).
Find the unknown parameter \(B\).
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x+5)(x+1)}{(x+5)(x+4)}\]
Notice, the factor \((x+5)\) is in both numerator and denominator. So, the hole is at \(x=-5\). Create a new function \(g\) with a filled in hole by eliminating common factors.
\[g(x)=\frac{x+1}{x+4}\]
Plug in \(x=-5\) into function \(g\).
\[g(-5)~=~\frac{-5+1}{-5+4}~=~\frac{-4}{-1}~=~4\]
Thus, the hole of \(y=f(x)\) is at point \((-5,4)\). This tells us that \(4=L(-5)\).
\[4={-5}+{B}\]
Subtract -5 from both sides.
\[9=B\]\[B=9\]
Notices this makes the line go through the hole.
You might find it interesting to use algebraic manipulations to solve \(f(x)=L(x)\), in order to see that you’d be led to an extraneous solution.
Which provides two solutions, \(x=-7\) and \(x=-5\). However, we know that \(x=-5\) cannot be a solution, because \(f(-5)\) is undefined! Thus, we see that if \(L(x)=x+{9}\), then we get an extraneous solution to \(f(x)=L(x)\).
This reinforces the implications of having \(B=9\), which makes the line go through the hole.
Let linear function \(L\) have a slope of 1 and an unknown parameter (\(b\)) dictating the \(y\)-intercept.
\[L(x)~=~x+B\]
Both curves, \(y=f(x)\) and \(y=L(x)\), are graphed. The hole of function \(f\) is at point \((h,k)\), and the point \((h,k)\) lies on the line, so \(k=L(h)\). This implies that the equation \(f(x)=L(x)\) would have an extraneous solution of \(x=h\).
Find the unknown parameter \(B\).
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x+7)(x-5)}{(x+10)(x+7)}\]
Notice, the factor \((x+7)\) is in both numerator and denominator. So, the hole is at \(x=-7\). Create a new function \(g\) with a filled in hole by eliminating common factors.
Which provides two solutions, \(x=-7\) and \(x=-5\). However, we know that \(x=-7\) cannot be a solution, because \(f(-7)\) is undefined! Thus, we see that if \(L(x)=x+{3}\), then we get an extraneous solution to \(f(x)=L(x)\).
This reinforces the implications of having \(B=3\), which makes the line go through the hole.
Question
Let rational function \(f\) be defined below.
\[f(x)~=~\frac{x^{2}-x-{56}}{x^{2}-{11}x+{24}}\]
Let linear function \(L\) have a slope of 1 and an unknown parameter (\(b\)) dictating the \(y\)-intercept.
\[L(x)~=~x+B\]
Both curves, \(y=f(x)\) and \(y=L(x)\), are graphed. The hole of function \(f\) is at point \((h,k)\), and the point \((h,k)\) lies on the line, so \(k=L(h)\). This implies that the equation \(f(x)=L(x)\) would have an extraneous solution of \(x=h\).
Find the unknown parameter \(B\).
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x+7)(x-8)}{(x-3)(x-8)}\]
Notice, the factor \((x-8)\) is in both numerator and denominator. So, the hole is at \(x=8\). Create a new function \(g\) with a filled in hole by eliminating common factors.
\[g(x)=\frac{x+7}{x-3}\]
Plug in \(x=8\) into function \(g\).
\[g(8)~=~\frac{8+7}{8-3}~=~\frac{15}{5}~=~3\]
Thus, the hole of \(y=f(x)\) is at point \((8,3)\). This tells us that \(3=L(8)\).
\[3={8}+{B}\]
Subtract 8 from both sides.
\[-5=B\]\[B=-5\]
Notices this makes the line go through the hole.
You might find it interesting to use algebraic manipulations to solve \(f(x)=L(x)\), in order to see that you’d be led to an extraneous solution.
Which provides two solutions, \(x=1\) and \(x=8\). However, we know that \(x=8\) cannot be a solution, because \(f(8)\) is undefined! Thus, we see that if \(L(x)=x-{5}\), then we get an extraneous solution to \(f(x)=L(x)\).
This reinforces the implications of having \(B=-5\), which makes the line go through the hole.
Question
Let rational function \(f\) be defined below.
\[f(x)~=~\frac{x^{2}+{6}x-{7}}{x^{2}-{3}x+{2}}\]
Let linear function \(L\) have a slope of 1 and an unknown parameter (\(b\)) dictating the \(y\)-intercept.
\[L(x)~=~x+B\]
Both curves, \(y=f(x)\) and \(y=L(x)\), are graphed. The hole of function \(f\) is at point \((h,k)\), and the point \((h,k)\) lies on the line, so \(k=L(h)\). This implies that the equation \(f(x)=L(x)\) would have an extraneous solution of \(x=h\).
Find the unknown parameter \(B\).
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x+7)(x-1)}{(x-1)(x-2)}\]
Notice, the factor \((x-1)\) is in both numerator and denominator. So, the hole is at \(x=1\). Create a new function \(g\) with a filled in hole by eliminating common factors.
\[g(x)=\frac{x+7}{x-2}\]
Plug in \(x=1\) into function \(g\).
\[g(1)~=~\frac{1+7}{1-2}~=~\frac{8}{-1}~=~-8\]
Thus, the hole of \(y=f(x)\) is at point \((1,-8)\). This tells us that \(-8=L(1)\).
\[-8={1}+{B}\]
Subtract 1 from both sides.
\[-9=B\]\[B=-9\]
Notices this makes the line go through the hole.
You might find it interesting to use algebraic manipulations to solve \(f(x)=L(x)\), in order to see that you’d be led to an extraneous solution.
Which provides two solutions, \(x=1\) and \(x=11\). However, we know that \(x=1\) cannot be a solution, because \(f(1)\) is undefined! Thus, we see that if \(L(x)=x-{9}\), then we get an extraneous solution to \(f(x)=L(x)\).
This reinforces the implications of having \(B=-9\), which makes the line go through the hole.
Question
Let rational function \(f\) be defined below.
\[f(x)~=~\frac{x^{2}+{7}x+{10}}{x^{2}+{3}x+{2}}\]
Let linear function \(L\) have a slope of 1 and an unknown parameter (\(b\)) dictating the \(y\)-intercept.
\[L(x)~=~x+B\]
Both curves, \(y=f(x)\) and \(y=L(x)\), are graphed. The hole of function \(f\) is at point \((h,k)\), and the point \((h,k)\) lies on the line, so \(k=L(h)\). This implies that the equation \(f(x)=L(x)\) would have an extraneous solution of \(x=h\).
Find the unknown parameter \(B\).
Solution
Factor the numerator, and factor the denominator.
\[f(x)~=~\frac{(x+5)(x+2)}{(x+2)(x+1)}\]
Notice, the factor \((x+2)\) is in both numerator and denominator. So, the hole is at \(x=-2\). Create a new function \(g\) with a filled in hole by eliminating common factors.
\[g(x)=\frac{x+5}{x+1}\]
Plug in \(x=-2\) into function \(g\).
\[g(-2)~=~\frac{-2+5}{-2+1}~=~\frac{3}{-1}~=~-3\]
Thus, the hole of \(y=f(x)\) is at point \((-2,-3)\). This tells us that \(-3=L(-2)\).
\[-3={-2}+{B}\]
Subtract -2 from both sides.
\[-1=B\]\[B=-1\]
Notices this makes the line go through the hole.
You might find it interesting to use algebraic manipulations to solve \(f(x)=L(x)\), in order to see that you’d be led to an extraneous solution.
Which provides two solutions, \(x=-2\) and \(x=3\). However, we know that \(x=-2\) cannot be a solution, because \(f(-2)\) is undefined! Thus, we see that if \(L(x)=x-{1}\), then we get an extraneous solution to \(f(x)=L(x)\).
This reinforces the implications of having \(B=-1\), which makes the line go through the hole.
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((102, 99)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-102)^2+(y_2-99)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(113,159)
2
(67,87)
3
(69,55)
4
(122,84)
5
(106,102)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-102)^2+(y_2-99)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(113-102)^2+(159-99)^2}~=~61\]
\[\sqrt{(67-102)^2+(87-99)^2}~=~37\]
\[\sqrt{(69-102)^2+(55-99)^2}~=~55\]
\[\sqrt{(122-102)^2+(84-99)^2}~=~25\]
\[\sqrt{(106-102)^2+(102-99)^2}~=~5\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((96, 103)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-96)^2+(y_2-103)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(111,123)
2
(156,148)
3
(85,43)
4
(112,73)
5
(66,63)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-96)^2+(y_2-103)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(111-96)^2+(123-103)^2}~=~25\]
\[\sqrt{(156-96)^2+(148-103)^2}~=~75\]
\[\sqrt{(85-96)^2+(43-103)^2}~=~61\]
\[\sqrt{(112-96)^2+(73-103)^2}~=~34\]
\[\sqrt{(66-96)^2+(63-103)^2}~=~50\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((105, 96)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-105)^2+(y_2-96)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(72,152)
2
(81,114)
3
(173,45)
4
(29,39)
5
(140,108)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-105)^2+(y_2-96)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(72-105)^2+(152-96)^2}~=~65\]
\[\sqrt{(81-105)^2+(114-96)^2}~=~30\]
\[\sqrt{(173-105)^2+(45-96)^2}~=~85\]
\[\sqrt{(29-105)^2+(39-96)^2}~=~95\]
\[\sqrt{(140-105)^2+(108-96)^2}~=~37\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((93, 106)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-93)^2+(y_2-106)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(177,71)
2
(77,118)
3
(73,154)
4
(75,26)
5
(28,34)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-93)^2+(y_2-106)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(177-93)^2+(71-106)^2}~=~91\]
\[\sqrt{(77-93)^2+(118-106)^2}~=~20\]
\[\sqrt{(73-93)^2+(154-106)^2}~=~52\]
\[\sqrt{(75-93)^2+(26-106)^2}~=~82\]
\[\sqrt{(28-93)^2+(34-106)^2}~=~97\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((105, 98)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-105)^2+(y_2-98)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(137,158)
2
(63,138)
3
(123,18)
4
(175,122)
5
(81,108)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-105)^2+(y_2-98)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(137-105)^2+(158-98)^2}~=~68\]
\[\sqrt{(63-105)^2+(138-98)^2}~=~58\]
\[\sqrt{(123-105)^2+(18-98)^2}~=~82\]
\[\sqrt{(175-105)^2+(122-98)^2}~=~74\]
\[\sqrt{(81-105)^2+(108-98)^2}~=~26\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((93, 97)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-93)^2+(y_2-97)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(153,177)
2
(72,125)
3
(165,43)
4
(25,148)
5
(17,40)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-93)^2+(y_2-97)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(153-93)^2+(177-97)^2}~=~100\]
\[\sqrt{(72-93)^2+(125-97)^2}~=~35\]
\[\sqrt{(165-93)^2+(43-97)^2}~=~90\]
\[\sqrt{(25-93)^2+(148-97)^2}~=~85\]
\[\sqrt{(17-93)^2+(40-97)^2}~=~95\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((101, 108)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-101)^2+(y_2-108)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(114,192)
2
(146,168)
3
(85,171)
4
(80,128)
5
(69,84)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-101)^2+(y_2-108)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(114-101)^2+(192-108)^2}~=~85\]
\[\sqrt{(146-101)^2+(168-108)^2}~=~75\]
\[\sqrt{(85-101)^2+(171-108)^2}~=~65\]
\[\sqrt{(80-101)^2+(128-108)^2}~=~29\]
\[\sqrt{(69-101)^2+(84-108)^2}~=~40\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((103, 106)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-103)^2+(y_2-106)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(148,78)
2
(121,186)
3
(68,22)
4
(67,121)
5
(58,166)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-103)^2+(y_2-106)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(148-103)^2+(78-106)^2}~=~53\]
\[\sqrt{(121-103)^2+(186-106)^2}~=~82\]
\[\sqrt{(68-103)^2+(22-106)^2}~=~91\]
\[\sqrt{(67-103)^2+(121-106)^2}~=~39\]
\[\sqrt{(58-103)^2+(166-106)^2}~=~75\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((103, 100)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-103)^2+(y_2-100)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(23,40)
2
(75,55)
3
(124,172)
4
(87,112)
5
(151,136)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-103)^2+(y_2-100)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(23-103)^2+(40-100)^2}~=~100\]
\[\sqrt{(75-103)^2+(55-100)^2}~=~53\]
\[\sqrt{(124-103)^2+(172-100)^2}~=~75\]
\[\sqrt{(87-103)^2+(112-100)^2}~=~20\]
\[\sqrt{(151-103)^2+(136-100)^2}~=~60\]
Question
The 2D distance formula is an equivalent equation to the Pythagorean equation. When finding distance between two points on an \(xy\)-plane, it can be helpful to imagine a right triangle with its legs horizontal and vertical, and its hypotenuse connecting the points. The length of the hypotenuse is the square root of the sum of the squares of the legs’ lengths. Each leg’s length is found by taking an absolute difference corresponding coordinates; for example \(|x_2-x_1|\) is the length of the horizontal leg, and \(|y_2-y_1|\) is the length of the vertical leg.
A square field is 200 feet by 200 feet. The field is marked with a grid pattern, so the Southwest corner is at point \((0,0)\) and the Northeast corner is at point \((200,200)\). Near the middle, at position \((95, 108)\) is a beacon, marked as a large bright-red dot. On the field are 5 players; their positions are listed in the table below. Find how far each player is from the beacon.
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-95)^2+(y_2-108)^2}\) and using sliders for \(x_2\) and \(y_2\).
Player
Position
Distance to beacon (ft)
1
(31,60)
2
(147,147)
3
(119,76)
4
(44,40)
5
(15,168)
Solution
I would recommend setting this up in Desmos using \(d=\sqrt{(x_2-95)^2+(y_2-108)^2}\) and using sliders for \(x_2\) and \(y_2\).
\[\sqrt{(31-95)^2+(60-108)^2}~=~80\]
\[\sqrt{(147-95)^2+(147-108)^2}~=~65\]
\[\sqrt{(119-95)^2+(76-108)^2}~=~40\]
\[\sqrt{(44-95)^2+(40-108)^2}~=~85\]
\[\sqrt{(15-95)^2+(168-108)^2}~=~100\]
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (93, 101).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (105, 97).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (102, 98).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (96, 107).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (91, 100).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (104, 97).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (103, 97).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (109, 99).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
Zoom out to see from 0 to 200 on both axes. Find where all three circles intersect by clicking.
All three circles intersect at point (97, 95).
Question
A circle has a center and a radius. The distance from the center to any point on the edge equals the radius.
In Desmos, we can graph a circle using an equation that looks like the Pythagorean equation and the 2D distance formula. If the center is at \((h,k)\) and the radius equals \(r\), the following equation graphs the desired circle.
\[(x-h)^2+(y-k)^2~=~r^2\]
A large square lake is 200 miles across. The positions of 3 lighthouses are known, as well as the distance from each lighthouse to a boat.
